MATHEMATICS

\displaystyle \text{Series BVM/1} \hspace{1.0cm} | \hspace{1.0cm} \text{Q. P. Code 65/1/3} \hspace{1.0cm} | \hspace{1.0cm} \text{Set 3 }    

\displaystyle \text{Time Allowed : 3 hours} \hspace{5.0cm} \text{Maximum Marks : 100 }  

\displaystyle \textbf{General Instructions :}
\displaystyle \text{(i) All questions are compulsory.}
\displaystyle \text{(ii) This question paper contains 29 questions divided into four sections A, B, C and D.}
\displaystyle \text{Section A comprises of 4 questions of one mark each, Section B comprises of } \\ \text{8 questions of two marks each,}   \text{Section C comprises of 11 questions of four marks} \\ \text{each and Section D comprises of 6 questions of six marks each.}
\displaystyle \text{(iii) All questions in Section A are to be answered in one word, one sentence or as } \\ \text{per the exact requirement of the question.}
\displaystyle \text{(iv) There is no overall choice. However, internal choice has been provided in } \\ \text{1 question of Section A, 3 questions of Section B,}
\displaystyle \text{3 questions of Section C and 3 questions of Section D. You have to attempt only } \\ \text{one of the alternatives in all such questions.}
\displaystyle \text{(v) Use of calculators is not permitted. You may ask logarithmic tables, if required.}

\displaystyle \textbf{SECTION - A}
\displaystyle \text{Question numbers 1 to 4 carry 1 mark each.}

\displaystyle \textbf{1. }\text{If A is a }3\times 3\text{ invertible matrix, then what will be the value }   \text{of }k\text{ if } \\ \det(A^{-1})=(\det A)^k.
\displaystyle \text{Answer:}
\displaystyle  \det(A^{-1})=\frac{1}{\det A}
\displaystyle \det(A^{-1})=(\det A)^k
\displaystyle \therefore (\det A)^{-1}=(\det A)^k
\displaystyle \therefore k=-1

\displaystyle \textbf{2. }\text{Determine the value of the constant }k\text{ so that the function}
\displaystyle f(x)=\begin{cases}\frac{kx}{|x|}, & x<0 \\ 3, & x\ge 0 \end{cases} \text{ is continuous at }x=0.
\displaystyle \text{Answer:}
\displaystyle  \lim_{x\to 0^-}f(x)=\lim_{x\to 0^-}\frac{kx}{|x|}=\lim_{x\to 0^-}\frac{kx}{-x}=-k
\displaystyle \lim_{x\to 0^+}f(x)=3
\displaystyle \text{As function is continuous at }x=0
\displaystyle \lim_{x\to 0^-}f(x)=\lim_{x\to 0^+}f(x)
\displaystyle 3=-k
\displaystyle \therefore k=-3

\displaystyle \textbf{3. }\text{Evaluate: }\int_{2}^{3} x^3\,dx.
\displaystyle \text{Answer:}
\displaystyle \text{Let }I=\int_{2}^{3}x^3dx
\displaystyle I=\left[\frac{x^4}{4}\right]_{2}^{3} =\frac{1}{4}(3^4-2^4) =\frac{1}{4}(81-16)=\frac{65}{4}

\displaystyle \textbf{4. }\text{If a line makes angles }90^\circ\text{ and }60^\circ\text{ respectively } \text{with the positive directions of }
\displaystyle x\text{ and }y\text{ axes, find the angle which it }   \text{makes with the positive direction of }z\text{-axis.}
\displaystyle \text{Answer:}
\displaystyle \text{Given }\alpha=90^\circ,\ \beta=60^\circ
\displaystyle \cos^2\alpha+\cos^2\beta+\cos^2\gamma=1
\displaystyle \cos^290^\circ+\cos^260^\circ+\cos^2\gamma=1
\displaystyle 0+\left(\frac{1}{2}\right)^2+\cos^2\gamma=1
\displaystyle \cos^2\gamma=\frac{3}{4}
\displaystyle \cos\gamma=\pm\frac{\sqrt{3}}{2}
\displaystyle \therefore \gamma=\frac{\pi}{6}\text{ or }\frac{5\pi}{6}

\displaystyle \textbf{SECTION - B}
\displaystyle \text{Question numbers 5 to 12 carry 2 marks each.}

\displaystyle \textbf{5. }\text{Show that all the diagonal elements of a skew symmetric }   \text{matrix are zero.}
\displaystyle \text{Answer:}
\displaystyle \text{Let }A=[a_{ij}]_{n\times n}\text{ be skew symmetric matrix}
\displaystyle A=-A^T
\displaystyle \Rightarrow a_{ij}=-a_{ji}\ \forall i,j
\displaystyle \text{For diagonal element }i=j
\displaystyle a_{ii}=-a_{ii}
\displaystyle \Rightarrow 2a_{ii}=0
\displaystyle \Rightarrow a_{ii}=0
\displaystyle \text{Therefore diagonal elements are zero.}

\displaystyle \textbf{6. }\text{Find }\frac{dy}{dx}\text{ at }x=1,y=\frac{\pi}{4}\text{ if }\sin^2y+\cos xy=K.
\displaystyle \text{Answer:}
\displaystyle \sin^2y+\cos(xy)=K
\displaystyle \text{Differentiating w.r.t }x
\displaystyle 2\sin y\cos y\frac{dy}{dx}-\sin(xy)\left(x\frac{dy}{dx}+y\right)=0
\displaystyle \sin2y\frac{dy}{dx}-x\sin(xy)\frac{dy}{dx}-y\sin(xy)=0
\displaystyle \frac{dy}{dx}(\sin2y-x\sin(xy))=y\sin(xy)
\displaystyle \frac{dy}{dx}=\frac{y\sin(xy)}{\sin2y-x\sin(xy)}

\displaystyle \textbf{7. }\text{The volume of a sphere is increasing at the rate of }3\text{ cubic centimeter per second.}
\displaystyle \text{Find the rate of increase of its surface}   \text{area, when the radius is }2\text{ cm.}
\displaystyle \text{Answer:}
\displaystyle V=\frac{4}{3}\pi r^3
\displaystyle \frac{dV}{dt}=4\pi r^2\frac{dr}{dt}
\displaystyle \frac{dV}{dt}=3
\displaystyle \therefore \frac{dr}{dt}=\frac{3}{4\pi r^2}
\displaystyle S=4\pi r^2
\displaystyle \frac{dS}{dt}=8\pi r\frac{dr}{dt}
\displaystyle \frac{dS}{dt}=8\pi r\left(\frac{3}{4\pi r^2}\right)=\frac{6}{r}
\displaystyle \text{At }r=2,\ \frac{dS}{dt}=3

\displaystyle \textbf{8. }\text{Show that the function }f(x)=4x^3-18x^2+27x-7\text{ is always }   \text{increasing on }R.
\displaystyle \text{Answer:}
\displaystyle f(x)=4x^3-18x^2+27x-7
\displaystyle f'(x)=12x^2-36x+27  =3(4x^2-12x+9)=3(2x-3)^2\ge 0\ \forall x\in R
\displaystyle \text{Therefore }f(x)\text{ is increasing on }R

\displaystyle \textbf{9. }\text{Find the vector equation of the line passing through the} \text{point }A(1,2,-1)
\displaystyle \text{ and parallel to the line }5x-25=14-7y=35z.
\displaystyle \text{Answer:}
\displaystyle \text{Given line }5x-25=14-7y=35z
\displaystyle \frac{x-5}{1/5}=\frac{y-2}{-1/7}=\frac{z}{1/35}
\displaystyle \text{Direction ratios }\left(\frac{1}{5},-\frac{1}{7},\frac{1}{35}\right)
\displaystyle \text{Required line passes through }(1,2,-1)
\displaystyle \overrightarrow{r}=(\widehat{i}+2\widehat{j}-\widehat{k})+\lambda\left(\frac{1}{5}\widehat{i}-\frac{1}{7}\widehat{j}+\frac{1}{35}\widehat{k}\right)

\displaystyle \textbf{10. }\text{Prove that if E and F are independent events, then the events } \text{E and }F'
\displaystyle \text{ are also independent.}
\displaystyle \text{Answer:}
\displaystyle P(E\cap F)=P(E)P(F)
\displaystyle P(E\cap F')=P(E)-P(E\cap F)
\displaystyle =P(E)-P(E)P(F)=P(E)[1-P(F)]=P(E)P(F')
\displaystyle \text{Therefore E and }F'\text{ are independent.}

\displaystyle \textbf{11. }\text{A small firm manufactures necklaces and bracelets. The total number of necklaces }
\displaystyle \text{and bracelets that it can handle per day is at It takes one hour to make a bracelet and half} \text{most}
\displaystyle \text{ 24 an hour to make a necklace. The maximum number of hours available per }\text{day is }16.
\displaystyle \text{ If the profit on a necklace is }\text{Rs }100\text{ and that on a} \text{bracelet is }\text{Rs }300, \text{ LLP for finding}
\displaystyle \text{how many of each should be produced daily to maximize the profit. It is being given }
\displaystyle \text{that at least one of each must be produced.}
\displaystyle \text{Answer:}
\displaystyle \text{Let number of necklaces be }x\text{ and bracelets be }y
\displaystyle \text{Maximum profit }P=100x+300y
\displaystyle x\ge 1,\ y\ge 1
\displaystyle x+y\le 24
\displaystyle \frac{1}{2}x+y\le 16\ \Rightarrow x+2y\le 32

\displaystyle \textbf{12. }\text{Find: }\int \frac{dx}{x^2+4x+8}.
\displaystyle \text{Answer:}
\displaystyle \text{Let }I=\int\frac{dx}{x^2+4x+8}
\displaystyle I=\int\frac{dx}{x^2+4x+4+4}
\displaystyle =\int\frac{dx}{(x+2)^2+2^2}
\displaystyle I=\frac{1}{2}\tan^{-1}\left(\frac{x+2}{2}\right)+C

\displaystyle \textbf{SECTION - C}
\displaystyle \text{Question numbers 13 to 23 carry 4 marks each.}

\displaystyle \textbf{13. }\text{Prove that }\tan\left(\frac{\pi}{4}+\frac{1}{2}\cos^{-1}\frac{a}{b}\right)   +\tan\left(\frac{\pi}{4}-\frac{1}{2}\cos^{-1}\frac{a}{b}\right)=\frac{2b}{a}.
\displaystyle \text{Answer:}
\displaystyle \text{Let }\frac{1}{2}\cos^{-1}\frac{a}{b}=x
\displaystyle \tan\left(\frac{\pi}{4}+x\right)+\tan\left(\frac{\pi}{4}-x\right)
\displaystyle =\frac{1+\tan x}{1-\tan x}+\frac{1-\tan x}{1+\tan x}
\displaystyle =\frac{(1+\tan x)^2+(1-\tan x)^2}{1-\tan^2 x}
\displaystyle =\frac{2(1+\tan^2 x)}{1-\tan^2 x}
\displaystyle =\frac{2(\sec^2 x)}{\cos^2 x-\sin^2 x}
\displaystyle =\frac{2}{\cos2x}
\displaystyle =\frac{2}{\cos\left(\cos^{-1}\frac{a}{b}\right)}
\displaystyle =\frac{2b}{a}

\displaystyle \textbf{14. }\text{Using properties of determinants, prove that}
\displaystyle \left|\begin{matrix}x & x+y & x+2y \\ x+2y & x & x+y \\ x+y & x+2y & x\end{matrix}\right|   =9y^2(x+y).
\displaystyle \text{OR}
\displaystyle \text{Let }A=\begin{bmatrix}2 & -1 \\ 3 & 4\end{bmatrix},B=\begin{bmatrix}5 & 2 \\ 7 & 4\end{bmatrix},   C=\begin{bmatrix}2 & 5 \\ 3 & 8\end{bmatrix},\text{ find a matrix }D\text{ such that } \\ CD-AB=0.
\displaystyle \text{Answer:}
\displaystyle \left|\begin{matrix}x & x+y & x+2y \\ x+2y & x & x+y \\ x+y & x+2y & x\end{matrix}\right|
\displaystyle \text{Apply }R_1\to R_1+R_2+R_3
\displaystyle =\left|\begin{matrix}3(x+y) & 3(x+y) & 3(x+y) \\ x+2y & x & x+y \\ x+y & x+2y & x\end{matrix}\right|
\displaystyle =3(x+y)\left|\begin{matrix}1 & 1 & 1 \\ x+2y & x & x+y \\ x+y & x+2y & x\end{matrix}\right|
\displaystyle \text{Apply }C_2\to C_2-C_1,\ C_3\to C_3-C_1
\displaystyle =3(x+y)\left|\begin{matrix}1 & 0 & 0 \\ x+2y & -2y & -y \\ x+y & y & -y\end{matrix}\right|
\displaystyle =3(x+y)\left|\begin{matrix}-2y & -y \\ y & -y\end{matrix}\right|
\displaystyle =3(x+y)(2y^2+y^2)
\displaystyle =9y^2(x+y)
\displaystyle \textbf{OR Let }D=\begin{bmatrix}x & y \\ z & w\end{bmatrix}
\displaystyle CD-AB=0\Rightarrow CD=AB
\displaystyle \begin{bmatrix}2 & 5 \\ 3 & 8\end{bmatrix}\begin{bmatrix}x & y \\ z & w\end{bmatrix}=\begin{bmatrix}2 & -1 \\ 3 & 4\end{bmatrix}\begin{bmatrix}5 & 2 \\ 7 & 4\end{bmatrix}
\displaystyle \begin{bmatrix}2x+5z & 2y+5w \\ 3x+8z & 3y+8w\end{bmatrix}=\begin{bmatrix}10-7 & 4-4 \\ 15+28 & 6+16\end{bmatrix}
\displaystyle \begin{bmatrix}2x+5z & 2y+5w \\ 3x+8z & 3y+8w\end{bmatrix}=\begin{bmatrix}3 & 0 \\ 43 & 22\end{bmatrix}
\displaystyle 2x+5z=3,\ 3x+8z=43
\displaystyle 2y+5w=0,\ 3y+8w=22
\displaystyle x=-191,\ z=77,\ y=-110,\ w=44
\displaystyle D=\begin{bmatrix}-191 & -110 \\ 77 & 44\end{bmatrix}

\displaystyle \textbf{15. }\text{Differentiate the function }(\sin x)^x+\sin^{-1}\sqrt{x}\text{ with respect to }x.
\displaystyle \text{OR}
\displaystyle \text{If }x^my^n=(x+y)^{m+n},\text{ prove that }\frac{d^2y}{dx^2}=0.
\displaystyle \text{Answer:}
\displaystyle \text{Let }y=(\sin x)^x+\sin^{-1}\sqrt{x}
\displaystyle y=u+v
\displaystyle \frac{dy}{dx}=\frac{du}{dx}+\frac{dv}{dx}
\displaystyle u=(\sin x)^x
\displaystyle \text{Taking log on both sides}
\displaystyle \log u=x\log(\sin x)
\displaystyle \text{Differentiating on both sides w.r.t. }x
\displaystyle \frac{1}{u}\frac{du}{dx}=x\frac{\cos x}{\sin x}+\log(\sin x)
\displaystyle \frac{du}{dx}=u[x\cot x+\log(\sin x)]
\displaystyle \frac{du}{dx}=(\sin x)^x[x\cot x+\log(\sin x)]
\displaystyle v=\sin^{-1}\sqrt{x}
\displaystyle \frac{dv}{dx}=\frac{1}{\sqrt{1-(\sqrt{x})^2}}\cdot\frac{1}{2\sqrt{x}}
\displaystyle \frac{dv}{dx}=\frac{1}{2\sqrt{x}\sqrt{1-x}}
\displaystyle =\frac{1}{2\sqrt{x-x^2}}
\displaystyle \therefore \frac{dy}{dx}=(\sin x)^x[x\cot x+\log(\sin x)]+\frac{1}{2\sqrt{x-x^2}}
\displaystyle \text{OR}
\displaystyle x^my^n=(x+y)^{m+n}
\displaystyle \text{Taking log on both sides}
\displaystyle \log(x^my^n)=\log(x+y)^{m+n}
\displaystyle \log x^m+\log y^n=\log(x+y)^{m+n}
\displaystyle m\log x+n\log y=(m+n)\log(x+y)
\displaystyle \text{Differentiating on both sides w.r.t. }x
\displaystyle \frac{m}{x}+\frac{n}{y}\frac{dy}{dx}=\frac{m+n}{x+y}\left(1+\frac{dy}{dx}\right)
\displaystyle \frac{n}{y}\frac{dy}{dx}-\frac{m+n}{x+y}\frac{dy}{dx}=\frac{m+n}{x+y}-\frac{m}{x}
\displaystyle \frac{dy}{dx}\left[\frac{nx+ny-my-ny}{y(x+y)}\right]=\frac{mx+nx-mx-my}{x(x+y)}
\displaystyle \frac{dy}{dx}\left[\frac{nx-my}{y(x+y)}\right]=\frac{nx-my}{x(x+y)}
\displaystyle \therefore \frac{dy}{dx}=\frac{y}{x}\qquad \cdots(1)
\displaystyle \text{Again differentiating w.r.t. }x
\displaystyle \frac{d^2y}{dx^2}=\frac{x\frac{dy}{dx}-y}{x^2}
\displaystyle =\frac{x\cdot\frac{y}{x}-y}{x^2}=0
\displaystyle \text{Hence proved.}

\displaystyle \textbf{16. }\text{Find: }\int \frac{2x}{(x^2+1)(x^2+2)^2}\,dx.
\displaystyle \text{Answer:}
\displaystyle \text{Let }I=\int\frac{2x}{(x^2+1)(x^2+2)^2}\,dx
\displaystyle \text{Let }x^2=y
\displaystyle 2x\,dx=dy
\displaystyle I=\int\frac{dy}{(y+1)(y+2)^2}
\displaystyle \text{Using partial fraction}
\displaystyle \frac{1}{(y+1)(y+2)^2}=\frac{A}{y+1}+\frac{B}{(y+2)^2}+\frac{C}{y+2}
\displaystyle 1=A(y+2)^2+B(y+1)+C(y+1)(y+2)
\displaystyle \text{When }y=-2,\ B=-1
\displaystyle \text{When }y=-1,\ A=1
\displaystyle \text{Put }y=0
\displaystyle 1=4-1+2C
\displaystyle C=-1
\displaystyle \therefore \frac{1}{(y+1)(y+2)^2}=\frac{1}{y+1}-\frac{1}{y+2}-\frac{1}{(y+2)^2}
\displaystyle I=\int\frac{dy}{y+1}-\int\frac{dy}{y+2}-\int\frac{dy}{(y+2)^2}
\displaystyle =\log(y+1)-\log(y+2)+\frac{1}{y+2}+C
\displaystyle \therefore I=\log(x^2+1)-\log(x^2+2)+\frac{1}{x^2+2}+C

\displaystyle \textbf{17. }\text{Evaluate: }\int_{0}^{\pi}\frac{x\sin x}{1+\cos^2x}\,dx.
\displaystyle \text{OR}
\displaystyle \text{Evaluate: }\int_{0}^{3/2}|x\sin \pi x|\,dx.
\displaystyle \text{Answer:}
\displaystyle \text{Let }I=\int_{0}^{\pi}\frac{x\sin x}{1+\cos^2x}\,dx
\displaystyle I=\int_{0}^{\pi}\frac{(\pi-x)\sin(\pi-x)}{1+\cos^2(\pi-x)}\,dx
\displaystyle I=\int_{0}^{\pi}\frac{\pi\sin x}{1+\cos^2x}\,dx-I
\displaystyle 2I=\int_{0}^{\pi}\frac{\pi\sin x}{1+\cos^2x}\,dx
\displaystyle \text{Let }\cos x=t,\ \text{then }-\sin x\,dx=dt
\displaystyle \text{When }x=0,\ t=1
\displaystyle \text{When }x=\pi,\ t=-1
\displaystyle 2I=-\pi\int_{1}^{-1}\frac{dt}{1+t^2}
\displaystyle 2I=\pi\int_{-1}^{1}\frac{dt}{1+t^2}
\displaystyle 2I=\pi\left[\tan^{-1}t\right]_{-1}^{1}
\displaystyle 2I=\pi\left(\frac{\pi}{4}+\frac{\pi}{4}\right)
\displaystyle 2I=\frac{\pi^2}{2}
\displaystyle I=\frac{\pi^2}{4}
\displaystyle \textbf{OR Let }I=\int_{0}^{3/2}|x\sin \pi x|\,dx
\displaystyle |x\sin \pi x|=\begin{cases}x\sin \pi x, & 0<x<1 \\ -x\sin \pi x, & 1<x<\frac{3}{2}\end{cases}
\displaystyle I=\int_{0}^{1}x\sin \pi x\,dx-\int_{1}^{3/2}x\sin \pi x\,dx
\displaystyle \text{Using integration by parts}
\displaystyle \int x\sin \pi x\,dx=-\frac{x\cos \pi x}{\pi}+\int\frac{\cos \pi x}{\pi}\,dx
\displaystyle =-\frac{x\cos \pi x}{\pi}+\frac{\sin \pi x}{\pi^2}
\displaystyle I=\left[-\frac{x\cos \pi x}{\pi}+\frac{\sin \pi x}{\pi^2}\right]_{0}^{1}-\left[-\frac{x\cos \pi x}{\pi}+\frac{\sin \pi x}{\pi^2}\right]_{1}^{3/2}
\displaystyle =\frac{2}{\pi}+\frac{1}{\pi^2}

\displaystyle \textbf{18. }\text{Prove that }x^2-y^2=C(x^2+y^2)^2\text{ is the general solution } \text{of the differential equation }
\displaystyle (x^3-3xy^2)dx=(y^3-3x^2y)dy,   \text{where }C\text{ is a parameter.}
\displaystyle \text{Answer:}
\displaystyle x^2-y^2=C(x^2+y^2)^2\qquad \cdots(1)
\displaystyle \text{Differentiating on both sides}
\displaystyle 2x-2y\frac{dy}{dx}=2C(x^2+y^2)\left(2x+2y\frac{dy}{dx}\right)
\displaystyle x-y\frac{dy}{dx}=C(x^2+y^2)\left(2x+2y\frac{dy}{dx}\right)\qquad \cdots(2)
\displaystyle \text{From }(1),\ C=\frac{x^2-y^2}{(x^2+y^2)^2}
\displaystyle \text{Putting value of }C\text{ in }(2)
\displaystyle x-y\frac{dy}{dx}=\frac{x^2-y^2}{x^2+y^2}\left(2x+2y\frac{dy}{dx}\right)
\displaystyle (x^2+y^2)\left(x-y\frac{dy}{dx}\right)=(x^2-y^2)\left(2x+2y\frac{dy}{dx}\right)
\displaystyle \frac{dy}{dx}\left[-2y(x^2-y^2)-y(x^2+y^2)\right]=2x(x^2-y^2)-x(y^2+x^2)
\displaystyle \frac{dy}{dx}\left(-3x^2y+y^3\right)=x^3-3xy^2
\displaystyle \frac{dy}{dx}(y^3-3x^2y)=x^3-3xy^2
\displaystyle (y^3-3x^2y)\,dy=(x^3-3xy^2)\,dx
\displaystyle \text{Hence }x^2-y^2=C(x^2+y^2)^2\text{ is the solution of the given}
\displaystyle \text{differential equation.}

\displaystyle \textbf{19. }\text{Let }\overrightarrow{a}=\widehat{i}+\widehat{j}+\widehat{k},\overrightarrow{b}=\widehat{i}\text{ and }   \overrightarrow{c}=c_1\widehat{i}+c_2\widehat{j}+c_3\widehat{k},\text{ then}
\displaystyle \text{(a) Let }c_1=1\text{ and }c_2=2,\text{ find }c_3\text{ which makes }\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}   \text{coplanar.}
\displaystyle \text{(b) If }c_2=-1\text{ and }c_3=1,\text{ show that no value of }c_1\text{ can make}   \overrightarrow{a},\overrightarrow{b},\overrightarrow{c}\text{ coplanar.}
\displaystyle \text{Answer:}
\displaystyle \overrightarrow{a}=\widehat{i}+\widehat{j}+\widehat{k}
\displaystyle \overrightarrow{b}=\widehat{i}
\displaystyle \overrightarrow{c}=c_1\widehat{i}+c_2\widehat{j}+c_3\widehat{k}
\displaystyle \overrightarrow{a},\overrightarrow{b},\overrightarrow{c}\text{ will be coplanar if}
\displaystyle [\overrightarrow{a}\ \overrightarrow{b}\ \overrightarrow{c}]=0
\displaystyle \left|\begin{matrix}1 & 1 & 1 \\ 1 & 0 & 0 \\ c_1 & c_2 & c_3\end{matrix}\right|=c_2-c_3
\displaystyle \text{(a) }c_1=1,\ c_2=2
\displaystyle [\overrightarrow{a}\ \overrightarrow{b}\ \overrightarrow{c}]=2-c_3
\displaystyle \text{As }\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}\text{ are coplanar}
\displaystyle 2-c_3=0
\displaystyle c_3=2
\displaystyle \text{(b) }c_2=-1,\ c_3=1
\displaystyle [\overrightarrow{a}\ \overrightarrow{b}\ \overrightarrow{c}]=c_2-c_3
\displaystyle =-1-1=-2\ne 0
\displaystyle \text{Therefore, no value of }c_1\text{ can make }\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}\text{ coplanar.}

\displaystyle \textbf{20. }\text{If }\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}\text{ are mutually perpendicular vectors of equal magnitudes, show that }
\displaystyle \text{the vector }\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}   \text{is equally inclined to }\overrightarrow{a},\overrightarrow{b}\text{ and }\overrightarrow{c}. \text{ Also, find the angle which}
\displaystyle \overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}\text{ makes with }\overrightarrow{a}\text{ or }\overrightarrow{b}\text{ or }\overrightarrow{c}.
\displaystyle \text{Answer:}
\displaystyle \text{Given: }|\overrightarrow{a}|=|\overrightarrow{b}|=|\overrightarrow{c}|\qquad \cdots(1)
\displaystyle \text{As vectors are perpendiculars}
\displaystyle \overrightarrow{a}\cdot\overrightarrow{b}=\overrightarrow{b}\cdot\overrightarrow{c}=\overrightarrow{c}\cdot\overrightarrow{a}=0\qquad \cdots(2)
\displaystyle \text{Let }\alpha,\beta,\gamma\text{ be the angles made by }(\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c})
\displaystyle \text{with }\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}\text{ respectively}
\displaystyle |\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}|^2=|\overrightarrow{a}|^2+|\overrightarrow{b}|^2+|\overrightarrow{c}|^2
\displaystyle \qquad +2(\overrightarrow{a}\cdot\overrightarrow{b}+\overrightarrow{b}\cdot\overrightarrow{c}+\overrightarrow{c}\cdot\overrightarrow{a})
\displaystyle |\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}|^2=3|\overrightarrow{a}|^2+2(0)\qquad \text{[From (1) and (2)]}
\displaystyle |\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}|^2=3|\overrightarrow{a}|^2\qquad \cdots(3)
\displaystyle (\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c})\cdot\overrightarrow{a}=|\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}||\overrightarrow{a}|\cos\alpha
\displaystyle \alpha=\cos^{-1}\left(\frac{|\overrightarrow{a}|}{|\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}|}\right)
\displaystyle \alpha=\cos^{-1}\left(\frac{1}{\sqrt{3}}\right)\qquad \text{[From (3)]}
\displaystyle \text{Similarly, }\beta=\cos^{-1}\left(\frac{1}{\sqrt{3}}\right)
\displaystyle \gamma=\cos^{-1}\left(\frac{1}{\sqrt{3}}\right)

\displaystyle \textbf{21. }\text{The random variable }X\text{ can take only the values }0,1,2,3.  \text{ Given that }
\displaystyle P(X=0)=P(X=1)=p\text{ and }P(X=2)=P(X=3)=q \text{ such that }
\displaystyle \sum p_ix_i^2=2\sum p_ix_i,\text{ find the value of }p.
\displaystyle \text{Answer:}
\displaystyle \text{Let }P(X=2)=P(X=3)=k
\displaystyle \begin{array}{|c|c|}\hline X & P(X) \\ \hline 0 & p \\ \hline 1 & p \\ \hline 2 & k \\ \hline 3 & k \\ \hline \end{array}
\displaystyle \sum P(X)=2p+2k
\displaystyle \text{Also }\sum P(X)=1
\displaystyle \therefore 2p+2k=1
\displaystyle 2k=1-2p
\displaystyle k=\frac{1}{2}-p
\displaystyle \begin{array}{|c|c|c|c|}\hline X_i & P_i & P_iX_i & P_iX_i^2 \\ \hline 0 & p & 0 & 0 \\ \hline 1 & p & p & p \\ \hline 2 & \frac{1}{2}-p & 1-2p & 2-4p \\ \hline 3 & \frac{1}{2}-p & \frac{3}{2}-3p & \frac{9}{2}-9p \\ \hline \end{array}
\displaystyle \sum P_iX_i=\frac{5}{2}-4p
\displaystyle \sum P_iX_i^2=\frac{13}{2}-13p
\displaystyle \sum P_iX_i^2=2\sum P_iX_i\qquad \text{(given)}
\displaystyle \frac{13}{2}-13p=2\left(\frac{5}{2}-4p\right)
\displaystyle \frac{13}{2}-\frac{10}{2}=13p-8p
\displaystyle \frac{3}{2}=5p
\displaystyle p=\frac{3}{10}

\displaystyle \textbf{22. }\text{Often it is taken that a truthful person commands more respect in the society. }
\displaystyle \text{A man is known to speak the truth }4\text{ out} \text{of }5 \text{ times. He throws a die and reports that }
\displaystyle \text{it is a six. Find the probability that it is actually a six. Do you also agree that the}
\displaystyle \text{value of truthfulness leads more respect in the society?}
\displaystyle \text{Answer:}
\displaystyle \text{Let }H_1\text{ be the event that }6\text{ appears on throwing a die}
\displaystyle H_2\text{ be the event that }6\text{ does not appear on throwing a die}
\displaystyle E\text{ be the event that he reports it is a six}
\displaystyle P(H_1)=\frac{1}{6}
\displaystyle P(H_2)=1-\frac{1}{6}=\frac{5}{6}
\displaystyle P(E\mid H_1)=\frac{4}{5}
\displaystyle P(E\mid H_2)=\frac{1}{5}
\displaystyle \text{Using Bayes theorem}
\displaystyle P(H_1\mid E)=\frac{P(H_1)\cdot P(E\mid H_1)}{P(H_1)\cdot P(E\mid H_1)+P(H_2)\cdot P(E\mid H_2)}
\displaystyle P(H_1\mid E)=\frac{\frac{1}{6}\times\frac{4}{5}}{\frac{1}{6}\times\frac{4}{5}+\frac{5}{6}\times\frac{1}{5}}
\displaystyle P(H_1\mid E)=\frac{4}{9}
\displaystyle \text{Yes, Truthfulness leads to more respect in society.}

\displaystyle \textbf{23. }\text{Solve the following L.P.P. graphically: Minimise }Z=5x+10y
\displaystyle \text{Subject to constraints }x+2y\le 120,\ x+y\ge 60,\ x-2y\ge 0\text{ and }x,y\ge 0.
\displaystyle \text{Answer:}  \displaystyle \text{Minimise }z=5x+10y
\displaystyle x+2y=120
\displaystyle \begin{array}{|c|c|c|c|}\hline x & 0 & 120 & 60 \\ \hline y & 60 & 0 & 30 \\ \hline \end{array}
\displaystyle x+y=60
\displaystyle \begin{array}{|c|c|c|c|}\hline x & 0 & 60 & 40 \\ \hline y & 60 & 0 & 20 \\ \hline \end{array}
\displaystyle x-2y=0
\displaystyle \begin{array}{|c|c|c|c|}\hline x & 0 & 60 & 40 \\ \hline y & 0 & 30 & 20 \\ \hline \end{array}
\displaystyle x\ge 0,\ y\ge 0
\displaystyle \begin{array}{|c|c|}\hline \text{Corner Points} & Z=5x+10y \\ \hline (60,0) & 300 \\ \hline (120,0) & 600 \\ \hline (40,20) & 400 \\ \hline (60,30) & 600 \\ \hline \end{array}
\displaystyle \text{Minimum value of }Z=300\text{ at }(60,0).

\displaystyle \textbf{SECTION - D}
\displaystyle \text{Question numbers 24 to 29 carry 6 marks each.}

\displaystyle \textbf{24. }\text{Use product }\begin{bmatrix}1 & -1 & 2 \\ 0 & 2 & -3 \\ 3 & -2 & 4\end{bmatrix}   \begin{bmatrix}-2 & 0 & 1 \\ 9 & 2 & -3 \\ 6 & 1 & -2\end{bmatrix}\text{ to solve the system of equations}
\displaystyle x+3z=9,\ -x+2y-2z=4,\ 2x-3y+4z=-3.
\displaystyle \text{Answer:}
\displaystyle \text{Let }A=\begin{bmatrix}1 & -1 & 2 \\ 0 & 2 & -3 \\ 3 & -2 & 4\end{bmatrix}
\displaystyle B=\begin{bmatrix}-2 & 0 & 1 \\ 9 & 2 & -3 \\ 6 & 1 & -2\end{bmatrix}
\displaystyle AB=\begin{bmatrix}1 & -1 & 2 \\ 0 & 2 & -3 \\ 3 & -2 & 4\end{bmatrix}\begin{bmatrix}-2 & 0 & 1 \\ 9 & 2 & -3 \\ 6 & 1 & -2\end{bmatrix}
\displaystyle =\begin{bmatrix}-2-9+12 & 0-2+2 & 1+3-4 \\ 0+18-18 & 0+4-3 & 0-6+6 \\ -6-18+24 & 0-4+4 & 3+6-8\end{bmatrix}
\displaystyle =\begin{bmatrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{bmatrix}=I
\displaystyle AB=I
\displaystyle B=A^{-1}I
\displaystyle B=A^{-1}
\displaystyle A^{-1}=\begin{bmatrix}-2 & 0 & 1 \\ 9 & 2 & -3 \\ 6 & 1 & -2\end{bmatrix}
\displaystyle \text{Given equations: }x+3z=9
\displaystyle -x+2y-2z=4
\displaystyle 2x-3y+4z=-3
\displaystyle \text{Matrix form}
\displaystyle \begin{bmatrix}1 & 0 & 3 \\ -1 & 2 & -2 \\ 2 & -3 & 4\end{bmatrix}\begin{bmatrix}x \\ y \\ z\end{bmatrix}=\begin{bmatrix}9 \\ 4 \\ -3\end{bmatrix}
\displaystyle A^TX=C
\displaystyle X=(A^T)^{-1}C
\displaystyle X=(A^{-1})^TC
\displaystyle \begin{bmatrix}x \\ y \\ z\end{bmatrix}=\begin{bmatrix}-2 & 9 & 6 \\ 0 & 2 & 1 \\ 1 & -3 & -2\end{bmatrix}\begin{bmatrix}9 \\ 4 \\ -3\end{bmatrix}
\displaystyle \begin{bmatrix}x \\ y \\ z\end{bmatrix}=\begin{bmatrix}-18+36-18 \\ 0+8-3 \\ 9-12+6\end{bmatrix}
\displaystyle \begin{bmatrix}x \\ y \\ z\end{bmatrix}=\begin{bmatrix}0 \\ 5 \\ 3\end{bmatrix}
\displaystyle \therefore x=0,\ y=5,\ z=3

\displaystyle \textbf{25. }\text{Consider }f:R^+\to[-5,\infty)\text{ given by }f(x)=9x^2+6x-5.  \text{ Show that } f \text{ is}
\displaystyle \text{invertible with }f^{-1}(y)=\frac{\sqrt{y+6}-1}{3}.
\displaystyle \text{Hence find: (i) }f^{-1}(10)\text{ (ii) }y\text{ if }f^{-1}(y)=\frac{4}{3}, \text{where }R^+ \text{ is the set of all non-negative}
\displaystyle \text{real numbers.}
\displaystyle \text{OR}
\displaystyle \text{Discuss the commutativity and associativity of binary operation *}  \ \text{defined on }
\displaystyle A=Q-\{1\}\text{ by }a*b=a-b+ab\text{ for all }a,b\in A.  \text{Also find the identity element of *}
\displaystyle \text{ in } A\text{ and hence find the}   \text{invertible elements of }A.
\displaystyle \text{Answer:}
\displaystyle \text{Clearly }f^{-1}(y)=g(y):[-5,\infty)\to R_+
\displaystyle f(x)=9x^2+6x-5
\displaystyle fog(y)=f\left(\frac{\sqrt{y+6}-1}{3}\right)  =9\left(\frac{\sqrt{y+6}-1}{3}\right)^2+6\left(\frac{\sqrt{y+6}-1}{3}\right)-5=y
\displaystyle \text{and }gof(x)=g(9x^2+6x-5)  =\frac{\sqrt{9x^2+6x+1}-1}{3}=x
\displaystyle \therefore g=f^{-1}
\displaystyle \therefore f\text{ is invertible}
\displaystyle \text{(i) }f^{-1}(10)=\frac{\sqrt{10+6}-1}{3}=\frac{\sqrt{16}-1}{3}=\frac{3}{3}=1
\displaystyle \text{(ii) }f^{-1}(y)=\frac{4}{3}
\displaystyle \frac{\sqrt{y+6}-1}{3}=\frac{4}{3}
\displaystyle \sqrt{y+6}=5
\displaystyle \text{Squaring on both sides}
\displaystyle y+6=25
\displaystyle y=19
\displaystyle \text{OR}
\displaystyle a*b=a-b+ab,\ \forall a,b\in A=Q-\{1\}
\displaystyle b*a=b-a+ba
\displaystyle a*b\ne b*a
\displaystyle \therefore *\text{ is not commutative}
\displaystyle (a*b)*c=(a-b+ab)*c =a-b-c+ab+ac-bc+abc
\displaystyle a*(b*c)=a*(b-c+bc)  =a-b+c+ab-bc+abc
\displaystyle (a*b)*c\ne a*(b*c)
\displaystyle \therefore *\text{ is not associative}
\displaystyle \text{Existence of identity}
\displaystyle a*e=a-e+ae=a
\displaystyle \Rightarrow e(a-1)=0
\displaystyle e=0
\displaystyle e*a=e-a+ea=a
\displaystyle e(1+a)=2a
\displaystyle e=\frac{2a}{1+a}
\displaystyle \therefore e\text{ is not unique}
\displaystyle \therefore \text{No identity element exists}
\displaystyle a*b=e=b*e
\displaystyle \therefore \text{No identity element exists}
\displaystyle \text{Inverse element does not exist}

\displaystyle \textbf{26. }\text{If the sum of lengths of the hypotenuse and a side of a right angled triangle is given,}
\displaystyle \text{show that the area of the triangle } \text{is maximum when the angle between them is }\frac{\pi}{3}.
\displaystyle \text{Answer:}
\displaystyle \text{Let base and hypotenuse of right angle triangle be }x\text{ and }y  \text{respectively}
\displaystyle \text{Also }\theta\text{ be angle between them}
\displaystyle \text{Given: Sum of length of side and hypotenous }
\displaystyle x+y=k
\displaystyle \text{Area of triangle }=\frac{1}{2}\times \text{Base}\times \text{height}\qquad \cdots(1)
\displaystyle AB^2=AC^2-BC^2\qquad \text{(By pythagoras theorem)}
\displaystyle AB=\sqrt{y^2-x^2}\qquad \cdots(2)
\displaystyle \therefore \text{Area of }\Delta=\frac{1}{2}x\sqrt{y^2-x^2}
\displaystyle A=\frac{1}{2}x\sqrt{(k-x)^2-x^2}\qquad \text{[}\because x+y=k\text{]}
\displaystyle A=\frac{1}{2}x\sqrt{k^2+x^2-2kx-x^2}
\displaystyle A=\frac{1}{2}x\sqrt{k^2-2kx}
\displaystyle \text{Squaring on both sides}
\displaystyle A^2=\frac{1}{4}x^2(k^2-2kx)
\displaystyle \text{Let }A^2=Z=\frac{1}{4}(k^2x^2-2kx^3)
\displaystyle \frac{dZ}{dx}=\frac{1}{4}(2k^2x-6kx^2)
\displaystyle \frac{dZ}{dx}=0
\displaystyle 2k^2x-6kx^2=0
\displaystyle 2k^2x=6kx^2
\displaystyle x=\frac{k}{3}\qquad \cdots(3)
\displaystyle x=\frac{x+y}{3}\qquad \text{[}\because k=x+y\text{]}
\displaystyle 3x=x+y
\displaystyle 2x=y\qquad \cdots(4)
\displaystyle \frac{d^2Z}{dx^2}=\frac{1}{4}(2k^2-12kx)
\displaystyle \left.\frac{d^2Z}{dx^2}\right|_{x=\frac{k}{3}}=\frac{1}{4}(2k^2-4k^2)=-\frac{k^2}{2}<0
\displaystyle \therefore \text{Area will be maximum for }2x=y
\displaystyle \cos\theta=\frac{x}{y}
\displaystyle \cos\theta=\frac{x}{2x}=\frac{1}{2}
\displaystyle \cos\theta=\cos\frac{\pi}{3}
\displaystyle \theta=\frac{\pi}{3}

\displaystyle \textbf{27. }\text{Using integration, find the area of region bounded by the } \text{triangle whose vertices are }
\displaystyle (-2,1),(0,4)\text{ and }(2,3).
\displaystyle \text{OR}
\displaystyle \text{Find the area bounded by the circle }x^2+y^2=16\text{ and the line } \sqrt{3}y=x \text{ in the first}
\displaystyle \text{quadrant, using integration.}
\displaystyle \text{Answer:}
\displaystyle \text{Vertices of triangle }(-2,1),(0,4)\text{ and }(2,3)\displaystyle \text{Equation of }AB:
\displaystyle y-y_1=\frac{y_2-y_1}{x_2-x_1}(x-x_1)
\displaystyle y-1=\frac{4-1}{0+2}(x+2)
\displaystyle y-1=\frac{3}{2}x+3
\displaystyle y=\frac{3}{2}x+4
\displaystyle \text{Similarly, equation of }BC:
\displaystyle y=4-\frac{x}{2}
\displaystyle \text{Equation of }AC:
\displaystyle y=\frac{1}{2}x+2
\displaystyle \text{Required area }=\int_{-2}^{0}(AB)\,dx+\int_{0}^{2}(BC)\,dx-\int_{-2}^{2}(AC)\,dx
\displaystyle =\int_{-2}^{0}\left(\frac{3}{2}x+4\right)dx+\int_{0}^{2}\left(4-\frac{x}{2}\right)dx-\int_{-2}^{2}\left(\frac{1}{2}x+2\right)dx
\displaystyle =\left(\frac{3}{4}x^2+4x\right)_{-2}^{0}+\left(4x-\frac{x^2}{4}\right)_{0}^{2}-\left(\frac{x^2}{4}+2x\right)_{-2}^{2}
\displaystyle =5+7-8=4\text{ sq. units}
\displaystyle \text{OR}
\displaystyle x^2+y^2=16
\displaystyle x=\sqrt{3}y\Rightarrow x^2=3y^2
\displaystyle 3y^2+y^2=16
\displaystyle y^2=4
\displaystyle y=\pm 2
\displaystyle x=\pm 2\sqrt{3}
\displaystyle \text{Point of intersection }(2\sqrt{3},2)\text{ and }(-2\sqrt{3},-2)
\displaystyle \text{Required area }=\int_{0}^{2\sqrt{3}}\frac{x}{\sqrt{3}}\,dx+\int_{2\sqrt{3}}^{4}\sqrt{4^2-x^2}\,dx
\displaystyle =\frac{1}{2\sqrt{3}}(x^2)_{0}^{2\sqrt{3}}+\left[\frac{x\sqrt{16-x^2}}{2}+8\sin^{-1}\frac{x}{4}\right]_{2\sqrt{3}}^{4}
\displaystyle =2\sqrt{3}+8\left(\frac{\pi}{2}-\frac{\pi}{3}\right)-2\sqrt{3}
\displaystyle =\frac{4\pi}{3}\text{ sq. units}
\displaystyle \text{Required area }=\sqrt{3}\int_{0}^{2}y\,dy+\int_{2}^{4}\sqrt{16-y^2}\,dy
\displaystyle =\sqrt{3}\left[\frac{y^2}{2}\right]_{0}^{2}+\left[\frac{y\sqrt{16-y^2}}{2}+8\sin^{-1}\frac{y}{4}\right]_{2}^{4}
\displaystyle =2\sqrt{3}+4\pi-2\sqrt{3}-\frac{4\pi}{3}=\frac{8\pi}{3}\text{ sq. units}

\displaystyle \textbf{28. }\text{Solve the differential equation }x\frac{dy}{dx}+y=x\cos x+\sin x,   \text{given that } \\ y=1\text{ when }x=\frac{\pi}{2}.
\displaystyle \text{Answer:}
\displaystyle x\frac{dy}{dx}+y=x\cos x+\sin x
\displaystyle \frac{dy}{dx}+\frac{y}{x}=\cos x+\frac{\sin x}{x}
\displaystyle \text{This is linear differential equation of the form }
\displaystyle \frac{dy}{dx}+Py=Q
\displaystyle \text{where }P=\frac{1}{x},\ Q=\cos x+\frac{\sin x}{x}
\displaystyle \therefore \text{Integrating factor}
\displaystyle I.F.=e^{\int Pdx}=e^{\int \frac{1}{x}dx}=e^{\log x}=x
\displaystyle \therefore \text{Solution is}
\displaystyle y\cdot x=\int\left(\cos x+\frac{\sin x}{x}\right)x\,dx+C
\displaystyle yx=\int x\cos x\,dx+\int \sin x\,dx+C
\displaystyle yx=x\sin x-\int \sin x\,dx+\int \sin x\,dx+C
\displaystyle yx=x\sin x+C
\displaystyle y=\sin x+\frac{C}{x}
\displaystyle \text{when }x=\frac{\pi}{2},\ y=1
\displaystyle 1=\sin\frac{\pi}{2}+\frac{C}{x}
\displaystyle C=0
\displaystyle \therefore \text{Particular solution is }y=\sin x

\displaystyle \textbf{29. }\text{Find the equation of the plane through the line of intersection of }
\displaystyle \overrightarrow{r}\cdot(2\widehat{i}-3\widehat{j}+4\widehat{k})=1\text{ and }   \overrightarrow{r}\cdot(\widehat{i}-\widehat{j})+4=0\text{ and perpendicular to the plane}
\displaystyle \overrightarrow{r}\cdot(2\widehat{i}-\widehat{j}+\widehat{k})+8=0.\text{ Hence find whether the plane thus obtained contains }
\displaystyle \text{the line }x-1=2y-4=3z-12.
\displaystyle \text{OR}
\displaystyle \text{Find the vector and Cartesian equations of a line passing through} (1,2,-4)
\displaystyle\text{ and perpendicular to the two lines }\frac{x-8}{3}=\frac{y+19}{-16}=\frac{z-10}{7} \text{ and }
\displaystyle \frac{x-15}{3}=\frac{y-29}{8}=\frac{z-5}{-5}.
\displaystyle \text{Answer:}
\displaystyle \text{The equation of the plane passing through the intersection of planes}
\displaystyle \vec{r}\cdot[(2\widehat{i}-\widehat{j}+\widehat{k})+\lambda(\widehat{i}-\widehat{j})]+4\lambda-1=0
\displaystyle \vec{r}\cdot[(2+\lambda)\widehat{i}-(3+\lambda)\widehat{j}+4\widehat{k}]+(4\lambda-1)=0\qquad \cdots(1)
\displaystyle \text{and it is perpendicular to the plane}
\displaystyle \vec{r}\cdot(2\widehat{i}-\widehat{j}+\widehat{k})+8=0
\displaystyle \therefore 2(2+\lambda)-1(-3-\lambda)+4(1)=0
\displaystyle 4+2\lambda+3+\lambda+4=0
\displaystyle 3\lambda=-11
\displaystyle \lambda=-\frac{11}{3}
\displaystyle \text{Putting value of }\lambda\text{ in (1)}
\displaystyle \vec{r}\cdot\left[\left(2-\frac{11}{3}\right)\widehat{i}-\left(3-\frac{11}{3}\right)\widehat{j}+4\widehat{k}\right]+\left[4\left(-\frac{11}{3}\right)-1\right]=0
\displaystyle \vec{r}\cdot\left(-\frac{5}{3}\widehat{i}+\frac{2}{3}\widehat{j}+4\widehat{k}\right)-\frac{47}{3}=0
\displaystyle \vec{r}\cdot(-5\widehat{i}+2\widehat{j}+12\widehat{k})=47
\displaystyle \text{Cartesian equation }-5x+2y+12z-47=0
\displaystyle \text{Line }\frac{x-1}{1}=\frac{y-2}{\frac{1}{2}}=\frac{z-4}{\frac{1}{3}}\text{ lies on plane}
\displaystyle \text{Point }(1,2,4)\text{ lies on plane}
\displaystyle (1,2,4)\cdot(-5,2,12)=47
\displaystyle -5+4+48=47
\displaystyle \text{Also }1(-5)+2\left(\frac{1}{2}\right)+12\left(\frac{1}{3}\right)=0
\displaystyle \therefore \text{plane contains given line}
\displaystyle \text{OR}
\displaystyle \text{Given lines:}
\displaystyle \frac{x-8}{3}=\frac{y+19}{-16}=\frac{z-10}{7}
\displaystyle \frac{x-15}{3}=\frac{y-29}{8}=\frac{z-5}{-5}
\displaystyle \text{Let }a,b,c\text{ be direction ratios}
\displaystyle 3a-16b+7c=0
\displaystyle 3a+8b-5c=0
\displaystyle \text{Solving: }\frac{a}{80-50}=\frac{b}{21+15}=\frac{c}{24+48}
\displaystyle \frac{a}{24}=\frac{b}{36}=\frac{c}{72}
\displaystyle \frac{a}{2}=\frac{b}{3}=\frac{c}{6}
\displaystyle \therefore \text{direction ratios }2,3,6
\displaystyle \text{Cartesian equation } \frac{x-1}{2}=\frac{y-2}{3}=\frac{z+4}{6}
\displaystyle \text{Vector equation }\vec{r}=(\widehat{i}+2\widehat{j}-4\widehat{k})+\lambda(2\widehat{i}+3\widehat{j}+6\widehat{k})

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