\displaystyle \textbf{Question 1: }~\text{Find the angle between the planes:}
\displaystyle \text{(i) }\ \overrightarrow{r}\cdot(2\widehat{i}-3\widehat{j}+4\widehat{k})=1\text{ and }\overrightarrow{r}\cdot(-\widehat{i}+\widehat{j})=4
\displaystyle \text{(ii) }\ \overrightarrow{r}\cdot(2\widehat{i}-\widehat{j}+2\widehat{k})=6\text{ and }\overrightarrow{r}\cdot(3\widehat{i}+6\widehat{j}-2\widehat{k})=9
\displaystyle \text{(iii) } \ \overrightarrow{r}\cdot(2\widehat{i}+3\widehat{j}-6\widehat{k})=5\text{ and }\overrightarrow{r}\cdot(\widehat{i}-2\widehat{j}+2\widehat{k})=9
\displaystyle \text{Answer:}
\displaystyle \text{(i)  }
\displaystyle \text{We know that the angle between the planes } \overrightarrow{r}\cdot\overrightarrow{n}_1=d_1,\ \overrightarrow{r}\cdot\overrightarrow{n}_2=d_2 \text{ is given by}
\displaystyle \cos\theta=\frac{\overrightarrow{n}_1\cdot\overrightarrow{n}_2}{\lvert\overrightarrow{n}_1\rvert\lvert\overrightarrow{n}_2\rvert}.
\displaystyle \text{Here, } \overrightarrow{n}_1=2\widehat{i}-3\widehat{j}+4\widehat{k},\ \overrightarrow{n}_2=-\widehat{i}+\widehat{j}+0\widehat{k}.
\displaystyle \text{So, } \cos\theta=\frac{(2\widehat{i}-3\widehat{j}+4\widehat{k})\cdot(-\widehat{i}+\widehat{j}+0\widehat{k})}{\lvert2\widehat{i}-3\widehat{j}+4\widehat{k}\rvert\lvert-\widehat{i}+\widehat{j}+0\widehat{k}\rvert}.
\displaystyle =\frac{-2-3+0}{\sqrt{4+9+16}\sqrt{1+1+0}}.
\displaystyle =\frac{-5}{\sqrt{29}\sqrt{2}}.
\displaystyle =\frac{-5}{\sqrt{58}}.
\displaystyle \Rightarrow \theta=\cos^{-1}\!\left(\frac{-5}{\sqrt{58}}\right).

\displaystyle \text{(ii)  }
\displaystyle \text{We know that the angle between the planes } \overrightarrow{r}\cdot\overrightarrow{n}_1=d_1,\ \overrightarrow{r}\cdot\overrightarrow{n}_2=d_2 \text{ is given by}
\displaystyle \cos\theta=\frac{\overrightarrow{n}_1\cdot\overrightarrow{n}_2}{\lvert\overrightarrow{n}_1\rvert\lvert\overrightarrow{n}_2\rvert}.
\displaystyle \text{Here, } \overrightarrow{n}_1=2\widehat{i}-\widehat{j}+2\widehat{k},\ \overrightarrow{n}_2=3\widehat{i}+6\widehat{j}-2\widehat{k}.
\displaystyle \text{So, } \cos\theta=\frac{(2\widehat{i}-\widehat{j}+2\widehat{k})\cdot(3\widehat{i}+6\widehat{j}-2\widehat{k})}{\lvert2\widehat{i}-\widehat{j}+2\widehat{k}\rvert\lvert3\widehat{i}+6\widehat{j}-2\widehat{k}\rvert}.
\displaystyle =\frac{6-6-4}{\sqrt{4+1+4}\sqrt{9+36+4}}.
\displaystyle =\frac{-4}{(3)(7)}.
\displaystyle =\frac{-4}{21}.
\displaystyle \Rightarrow \theta=\cos^{-1}\!\left(\frac{-4}{21}\right).

\displaystyle \text{(iii)  }
\displaystyle \text{We know that the angle between the planes } \overrightarrow{r}\cdot\overrightarrow{n}_1=d_1,\ \overrightarrow{r}\cdot\overrightarrow{n}_2=d_2 \text{ is given by}
\displaystyle \cos\theta=\frac{\overrightarrow{n}_1\cdot\overrightarrow{n}_2}{\lvert\overrightarrow{n}_1\rvert\lvert\overrightarrow{n}_2\rvert}.
\displaystyle \text{Here, } \overrightarrow{n}_1=2\widehat{i}+3\widehat{j}-6\widehat{k},\ \overrightarrow{n}_2=\widehat{i}-2\widehat{j}+2\widehat{k}.
\displaystyle \text{So, } \cos\theta=\frac{(2\widehat{i}+3\widehat{j}-6\widehat{k})\cdot(\widehat{i}-2\widehat{j}+2\widehat{k})}{\lvert2\widehat{i}+3\widehat{j}-6\widehat{k}\rvert\lvert\widehat{i}-2\widehat{j}+2\widehat{k}\rvert}.
\displaystyle =\frac{2-6-12}{\sqrt{4+9+36}\sqrt{1+4+4}}.
\displaystyle =\frac{-16}{(7)(3)}.
\displaystyle =\frac{-16}{21}.
\displaystyle \Rightarrow \theta=\cos^{-1}\!\left(\frac{-16}{21}\right).

\displaystyle \textbf{Question 2: }~\text{Find the angle between the planes:}
\displaystyle \text{(i) }\ 2x-y+z=4\text{ and }x+y+2z=3
\displaystyle \text{(ii) }\ x+y-2z=3\text{ and }2x-2y+z=5
\displaystyle \text{(iii) }\ x-y+z=5\text{ and }x+2y+z=9
\displaystyle\text{ (iv) }\ 2x-3y+4z=1\text{ and }-x+y=4
\displaystyle \text{(v) }\ 2x+y-2z=5\text{ and }3x-6y-2z=7 
\displaystyle \text{Answer:}
\displaystyle \text{(i)  }
\displaystyle \text{We know that the angle between the planes } a_1x+b_1y+c_1z+d_1=0 \text{ and } a_2x+b_2y+c_2z+d_2=0 \text{ is given by}
\displaystyle \cos\theta=\frac{a_1a_2+b_1b_2+c_1c_2}{\sqrt{a_1^2+b_1^2+c_1^2}\sqrt{a_2^2+b_2^2+c_2^2}}.
\displaystyle \text{So, the angle between } 2x-y+z=4 \text{ and } x+y+2z=3 \text{ is given by}
\displaystyle \cos\theta=\frac{(2)(1)+(-1)(1)+(1)(2)}{\sqrt{2^2+(-1)^2+1^2}\sqrt{1^2+1^2+2^2}}.
\displaystyle =\frac{2-1+2}{\sqrt{4+1+1}\sqrt{1+1+4}}.
\displaystyle =\frac{3}{\sqrt{6}\sqrt{6}}.
\displaystyle =\frac{3}{6}.
\displaystyle =\frac{1}{2}.
\displaystyle \Rightarrow \theta=\cos^{-1}\!\left(\frac{1}{2}\right)=\frac{\pi}{3}.

\displaystyle \text{(ii)  }
\displaystyle \text{We know that the angle between the planes } a_1x+b_1y+c_1z+d_1=0 \text{ and } a_2x+b_2y+c_2z+d_2=0 \text{ is given by}
\displaystyle \cos\theta=\frac{a_1a_2+b_1b_2+c_1c_2}{\sqrt{a_1^2+b_1^2+c_1^2}\sqrt{a_2^2+b_2^2+c_2^2}}.
\displaystyle \text{So, the angle between } x+y-2z=3 \text{ and } 2x-2y+z=5 \text{ is given by}
\displaystyle \cos\theta=\frac{(1)(2)+(1)(-2)+(-2)(1)}{\sqrt{1^2+1^2+(-2)^2}\sqrt{2^2+(-2)^2+1^2}}.
\displaystyle =\frac{2-2-2}{\sqrt{1+1+4}\sqrt{4+4+1}}.
\displaystyle =\frac{-2}{\sqrt{6}\sqrt{9}}.
\displaystyle =\frac{-2}{3\sqrt{6}}.
\displaystyle \Rightarrow \theta=\cos^{-1}\!\left(\frac{-2}{3\sqrt{6}}\right).

\displaystyle \text{(iii)  }
\displaystyle \text{We know that the angle between the planes } a_1x+b_1y+c_1z+d_1=0 \text{ and } a_2x+b_2y+c_2z+d_2=0 \text{ is given by}
\displaystyle \cos\theta=\frac{a_1a_2+b_1b_2+c_1c_2}{\sqrt{a_1^2+b_1^2+c_1^2}\sqrt{a_2^2+b_2^2+c_2^2}}.
\displaystyle \text{So, the angle between } x-y+z=5 \text{ and } x+2y+z=9 \text{ is given by}
\displaystyle \cos\theta=\frac{(1)(1)+(-1)(2)+(1)(1)}{\sqrt{1^2+(-1)^2+1^2}\sqrt{1^2+2^2+1^2}}.
\displaystyle =\frac{1-2+1}{\sqrt{1+1+1}\sqrt{1+4+1}}.
\displaystyle =\frac{0}{\sqrt{3}\sqrt{6}}.
\displaystyle =0.
\displaystyle \Rightarrow \theta=\cos^{-1}(0)=\frac{\pi}{2}.

\displaystyle \text{(iv)  }
\displaystyle \text{We know that the angle between the planes } a_1x+b_1y+c_1z+d_1=0 \text{ and } a_2x+b_2y+c_2z+d_2=0 \text{ is given by}
\displaystyle \cos\theta=\frac{a_1a_2+b_1b_2+c_1c_2}{\sqrt{a_1^2+b_1^2+c_1^2}\sqrt{a_2^2+b_2^2+c_2^2}}.
\displaystyle \text{So, the angle between } 2x-3y+4z=1 \text{ and } -x+y+0z=4 \text{ is given by}
\displaystyle \cos\theta=\frac{(2)(-1)+(-3)(1)+(4)(0)}{\sqrt{2^2+(-3)^2+4^2}\sqrt{(-1)^2+1^2+0^2}}.
\displaystyle =\frac{-2-3+0}{\sqrt{4+9+16}\sqrt{1+1+0}}.
\displaystyle =\frac{-5}{\sqrt{29}\sqrt{2}}.
\displaystyle =\frac{-5}{\sqrt{58}}.
\displaystyle \Rightarrow \theta=\cos^{-1}\!\left(\frac{-5}{\sqrt{58}}\right).

\displaystyle \text{(v)  }
\displaystyle \text{We know that the angle between the planes } a_1x+b_1y+c_1z+d_1=0 \text{ and } a_2x+b_2y+c_2z+d_2=0 \text{ is given by}
\displaystyle \cos\theta=\frac{a_1a_2+b_1b_2+c_1c_2}{\sqrt{a_1^2+b_1^2+c_1^2}\sqrt{a_2^2+b_2^2+c_2^2}}.
\displaystyle \text{So, the angle between } 2x+y-2z=5 \text{ and } 3x-6y-2z=7 \text{ is given by}
\displaystyle \cos\theta=\frac{(2)(3)+(1)(-6)+(-2)(-2)}{\sqrt{2^2+1^2+(-2)^2}\sqrt{3^2+(-6)^2+(-2)^2}}.
\displaystyle =\frac{6-6+4}{\sqrt{4+1+4}\sqrt{9+36+4}}.
\displaystyle =\frac{4}{(3)(7)}.
\displaystyle =\frac{4}{21}.
\displaystyle \Rightarrow \theta=\cos^{-1}\!\left(\frac{4}{21}\right).

\displaystyle \textbf{Question 3: }~\text{Show that the following planes are at right angles:}
\displaystyle \text{(i) } \overrightarrow{r}\cdot(2\widehat{i}-\widehat{j}+\widehat{k})=5\text{ and }\overrightarrow{r}\cdot(-\widehat{i}-\widehat{j}+\widehat{k})=3
\displaystyle \text{(ii) }\ x-2y+4z=10\text{ and }18x+17y+4z=49
\displaystyle \text{Answer:}
\displaystyle \text{(i)  }
\displaystyle \text{We know that the planes } \overrightarrow{r}\cdot\overrightarrow{n}_1=d_1,\ \overrightarrow{r}\cdot\overrightarrow{n}_2=d_2 \text{ are perpendicular to each other only if } \overrightarrow{n}_1\cdot\overrightarrow{n}_2=0.
\displaystyle \text{Here, } \overrightarrow{n}_1=2\widehat{i}-\widehat{j}+\widehat{k},\ \overrightarrow{n}_2=-\widehat{i}-\widehat{j}+\widehat{k}.
\displaystyle \text{Now,}
\displaystyle \overrightarrow{n}_1\cdot\overrightarrow{n}_2=(2\widehat{i}-\widehat{j}+\widehat{k})\cdot(-\widehat{i}-\widehat{j}+\widehat{k})=-2+1+1.
\displaystyle =0.
\displaystyle \text{So, the given planes are perpendicular.}

\displaystyle \text{(ii)  }
\displaystyle \text{We know that the planes } a_1x+b_1y+c_1z+d_1=0 \text{ and } a_2x+b_2y+c_2z+d_2=0 \text{ are perpendicular to each other only if}
\displaystyle a_1a_2+b_1b_2+c_1c_2=0.
\displaystyle \text{The given planes are } x-2y+4z=10 \text{ and } 18x+17y+4z=49.
\displaystyle \Rightarrow a_1=1,\ b_1=-2,\ c_1=4,\ a_2=18,\ b_2=17,\ c_2=4.
\displaystyle \text{Now,}
\displaystyle a_1a_2+b_1b_2+c_1c_2=(1)(18)+(-2)(17)+(4)(4)=18-34+16.
\displaystyle =0.
\displaystyle \text{So, the given planes are perpendicular.}

\displaystyle \textbf{Question 4: }~\text{Determine the value of }\lambda\text{ for which the following planes are perpendicular to each other.}
\displaystyle \text{(i) }\ \overrightarrow{r}\cdot(\widehat{i}+2\widehat{j}+3\widehat{k})=7\text{ and }\overrightarrow{r}\cdot(\lambda\widehat{i}+2\widehat{j}-7\widehat{k})=26
\displaystyle \text{(ii) }\ 2x-4y+3z=5\text{ and }x+2y+\lambda z=5
\displaystyle \text{(iii) }\ 3x-6y-2z=7\text{ and }2x+y-\lambda z=5
\displaystyle \text{Answer:}
\displaystyle \text{(i)  }
\displaystyle \text{We know that the planes } \overrightarrow{r}\cdot\overrightarrow{n}_1=d_1,\ \overrightarrow{r}\cdot\overrightarrow{n}_2=d_2 \text{ are perpendicular to each other only if } \overrightarrow{n}_1\cdot\overrightarrow{n}_2=0.
\displaystyle \text{Here, } \overrightarrow{n}_1=\widehat{i}+2\widehat{j}+3\widehat{k},\ \overrightarrow{n}_2=\lambda\widehat{i}+2\widehat{j}-7\widehat{k}.
\displaystyle \text{The given planes are perpendicular.}
\displaystyle \Rightarrow \overrightarrow{n}_1\cdot\overrightarrow{n}_2=0.
\displaystyle \Rightarrow (\widehat{i}+2\widehat{j}+3\widehat{k})\cdot(\lambda\widehat{i}+2\widehat{j}-7\widehat{k})=0.
\displaystyle \Rightarrow \lambda+4-21=0.
\displaystyle \Rightarrow \lambda-17=0.
\displaystyle \Rightarrow \lambda=17.

\displaystyle \text{(ii)  }
\displaystyle \text{We know that the planes } a_1x+b_1y+c_1z+d_1=0 \text{ and } a_2x+b_2y+c_2z+d_2=0 \text{ are perpendicular to each other only if}
\displaystyle a_1a_2+b_1b_2+c_1c_2=0.
\displaystyle \text{The given planes are } 2x-4y+3z=5 \text{ and } x+2y+\lambda z=5.
\displaystyle \Rightarrow a_1=2,\ b_1=-4,\ c_1=3,\ a_2=1,\ b_2=2,\ c_2=\lambda.
\displaystyle \text{It is given that the planes are perpendicular.}
\displaystyle \Rightarrow a_1a_2+b_1b_2+c_1c_2=0.
\displaystyle \Rightarrow (2)(1)+(-4)(2)+(3)(\lambda)=0.
\displaystyle \Rightarrow 2-8+3\lambda=0.
\displaystyle \Rightarrow 3\lambda=6.
\displaystyle \Rightarrow \lambda=2.

\displaystyle \text{(iii)  }
\displaystyle \text{We know that the planes } a_1x+b_1y+c_1z+d_1=0 \text{ and } a_2x+b_2y+c_2z+d_2=0 \text{ are perpendicular to each other only if}
\displaystyle a_1a_2+b_1b_2+c_1c_2=0.
\displaystyle \text{The given planes are } 3x-6y-2z=7 \text{ and } 2x+y-\lambda z=5.
\displaystyle \Rightarrow a_1=3,\ b_1=-6,\ c_1=-2,\ a_2=2,\ b_2=1,\ c_2=-\lambda.
\displaystyle \text{The given planes are perpendicular.}
\displaystyle \Rightarrow a_1a_2+b_1b_2+c_1c_2=0.
\displaystyle \Rightarrow (3)(2)+(-6)(1)+(-2)(-\lambda)=0.
\displaystyle \Rightarrow 6-6+2\lambda=0.
\displaystyle \Rightarrow 2\lambda=0.
\displaystyle \Rightarrow \lambda=0.

\displaystyle \textbf{Question 5: }~\text{Find the equation of a plane passing through the point } \\ (-1,-1,2)\text{ and perpendicular to the planes }3x+2y-3z=1\text{ and } \\ 5x-4y+z=5.\ \hspace{7.0cm} \text{[CBSE\ 2004,\ 2008]}
\displaystyle \text{Answer:}
\displaystyle \text{The equation of any plane passing through }(-1,-1,2)\text{ is}
\displaystyle a(x+1)+b(y+1)+c(z-2)=0.\qquad(1)
\displaystyle \text{It is given that (1) is perpendicular to each of the planes }3x+2y-3z=1 \text{ and } 5x-4y+z=5.\ \text{Then,}
\displaystyle 3a+2b-3c=0.\qquad(2)
\displaystyle 5a-4b+c=0.\qquad(3)
\displaystyle \text{Solving (1), (2) and (3), we get}
\displaystyle \begin{vmatrix}x+1&y+1&z-2\\3&2&-3\\5&-4&1\end{vmatrix}=0.
\displaystyle \Rightarrow -10(x+1)-18(y+1)-22(z-2)=0.
\displaystyle \Rightarrow 5(x+1)+9(y+1)+11(z-2)=0.
\displaystyle \Rightarrow 5x+5+9y+9+11z-22=0.
\displaystyle \Rightarrow 5x+9y+11z-8=0.

\displaystyle \textbf{Question 6: }~\text{Obtain the equation of the plane passing through the point } \\ (1,-3,-2)\text{ and perpendicular to the planes }x+2y+2z=5\text{ and } \\ 3x+3y+2z=8.\ \hspace{7.0cm} \text{[CBSE\ 2009]}
\displaystyle \text{Answer:}
\displaystyle \text{The equation of any plane passing through }(1,-3,-2)\text{ is}
\displaystyle a(x-1)+b(y+3)+c(z+2)=0.\qquad(1)
\displaystyle \text{It is given that (1) is perpendicular to the planes } x+2y+2z=5 \text{ and } 3x+3y+2z=8.\ \text{Then,}
\displaystyle a+2b+2c=0.\qquad(2)
\displaystyle 3a+3b+2c=0.\qquad(3)
\displaystyle \text{Solving (1), (2) and (3), we get}
\displaystyle \begin{vmatrix}x-1&y+3&z+2\\1&2&2\\3&3&2\end{vmatrix}=0.
\displaystyle \Rightarrow -2(x-1)+4(y+3)-3(z+2)=0.
\displaystyle \Rightarrow -2x+2+4y+12-3z-6=0.
\displaystyle \Rightarrow 2x-4y+3z-8=0.

\displaystyle \textbf{Question 7: }~\text{Find the equation of the plane passing through the origin and} \\ \text{perpendicular to each of the planes }x+2y-z=1\text{ and } 3x-4y+z=5.
\displaystyle \text{Answer:}
\displaystyle \text{The equation of any plane passing through the origin }(0,0,0)\text{ is}
\displaystyle a(x-0)+b(y-0)+c(z-0)=0.
\displaystyle ax+by+cz=0.\qquad(1)
\displaystyle \text{It is given that (1) is perpendicular to the planes } x+2y-z=1 \text{ and } 3x-4y+z=5.\ \text{Then,}
\displaystyle a+2b-c=0.\qquad(2)
\displaystyle 3a-4b+c=0.\qquad(3)
\displaystyle \text{Solving (1), (2) and (3), we get}
\displaystyle \begin{vmatrix}x&y&z\\1&2&-1\\3&-4&1\end{vmatrix}=0.
\displaystyle \Rightarrow -2x-4y-10z=0.
\displaystyle \Rightarrow x+2y+5z=0.

\displaystyle \textbf{Question 8: }~\text{Find the equation of the plane passing through the points } \\ (1,-1,2)\text{ and }(2,-2,2)\text{ and which is perpendicular to the plane } \\ 6x-2y+2z=9.\  \hspace{7.0cm} \text{[CBSE\ 2005]}
\displaystyle \text{Answer:}
\displaystyle \text{The equation of any plane passing through }(1,-1,2)\text{ is}
\displaystyle a(x-1)+b(y+1)+c(z-2)=0.\qquad(1)
\displaystyle \text{It is given that (1) is passing through }(2,-2,2).\ \text{So,}
\displaystyle a(2-1)+b(-2+1)+c(2-2)=0.
\displaystyle \Rightarrow a-b+0c=0.\qquad(2)
\displaystyle \text{It is given that (1) is perpendicular to the plane }6x-2y+2z=9.\ \text{So,}
\displaystyle 6a-2b+2c=0.
\displaystyle \Rightarrow 3a-b+c=0.\qquad(3)
\displaystyle \text{Solving (1), (2) and (3), we get}
\displaystyle \begin{vmatrix}x-1&y+1&z-2\\1&-1&0\\3&-1&1\end{vmatrix}=0.
\displaystyle \Rightarrow -1(x-1)-1(y+1)+2(z-2)=0.
\displaystyle \Rightarrow -x+1-y-1+2z-4=0.
\displaystyle \Rightarrow x+y-2z+4=0.

\displaystyle \textbf{Question 9: }~\text{Find the equation of the plane passing through the points } \\ (2,2,1)\text{ and }(9,3,6)\text{ and perpendicular to the plane }2x+6y+6z=1.
\displaystyle \text{Answer:}
\displaystyle \text{The equation of any plane passing through }(2,2,1)\text{ is}
\displaystyle a(x-2)+b(y-2)+c(z-1)=0.\qquad(1)
\displaystyle \text{It is given that (1) is passing through }(9,3,6).\ \text{So,}
\displaystyle a(9-2)+b(3-2)+c(6-1)=0.
\displaystyle \Rightarrow 7a+b+5c=0.\qquad(2)
\displaystyle \text{It is given that (1) is perpendicular to the plane }2x+6y+6z=1.\ \text{So,}
\displaystyle 2a+6b+6c=0.
\displaystyle \Rightarrow a+3b+3c=0.\qquad(3)
\displaystyle \text{Solving (1), (2) and (3), we get}
\displaystyle \begin{vmatrix}x-2&y-2&z-1\\7&1&5\\1&3&3\end{vmatrix}=0.
\displaystyle \Rightarrow -12(x-2)-16(y-2)+20(z-1)=0.
\displaystyle \Rightarrow 3(x-2)+4(y-2)-5(z-1)=0.
\displaystyle \Rightarrow 3x+4y-5z=9.

\displaystyle \textbf{Question 10: }~\text{Find the equation of the plane passing through the points whose} \\ \text{coordinates are }(-1,1,1)\text{ and }(1,-1,1)\text{ and perpendicular to the plane } \\ x+2y+2z=5.
\displaystyle \text{Answer:}
\displaystyle \text{The equation of any plane passing through }(-1,1,1)\text{ is}
\displaystyle a(x+1)+b(y-1)+c(z-1)=0.\qquad(1)
\displaystyle \text{It is given that (1) is passing through }(1,-1,1).\ \text{So,}
\displaystyle a(1+1)+b(-1-1)+c(1-1)=0.
\displaystyle \Rightarrow 2a-2b+0c=0.\qquad(2)
\displaystyle \text{It is given that (1) is perpendicular to the plane }x+2y+2z=5.\ \text{So,}
\displaystyle a+2b+2c=0.\qquad(3)
\displaystyle \text{Solving (1), (2) and (3), we get}
\displaystyle \begin{vmatrix}x+1&y-1&z-1\\2&-2&0\\1&2&2\end{vmatrix}=0.
\displaystyle \Rightarrow -4(x+1)-4(y-1)+6(z-1)=0.
\displaystyle \Rightarrow 2(x+1)+2(y-1)-3(z-1)=0.
\displaystyle \Rightarrow 2x+2y-3z+3=0.

\displaystyle \textbf{Question 11: }~\text{Find the equation of the plane with intercept }3\text{ on the } \\ y\text{-axis and parallel to }ZOX\text{ plane.}
\displaystyle \text{Answer:}
\displaystyle \text{The equation of the plane parallel to the plane ZOX is}
\displaystyle y=b.\qquad(1)
\displaystyle \text{where } b \text{ is a constant.}
\displaystyle \text{It is given that this plane passes through }(0,3,0).\ \text{So,}
\displaystyle 3=b.
\displaystyle \text{Substituting this value in (1), we get}
\displaystyle y=3,\ \text{which is the required equation of the plane.}

\displaystyle \textbf{Question 12: }~\text{Find the equation of the plane that contains the point } \\ (1,-1,2)\text{ and is perpendicular to each of the planes }2x+3y-2z=5\text{ and } \\ x+2y-3z=8.\ \hspace{7.0cm}  \text{[CBSE\ 2014 ]}
\displaystyle \text{Answer:}
\displaystyle \text{The equation of any plane passing through }(1,-1,2)\text{ is}
\displaystyle a(x-1)+b(y+1)+c(z-2)=0.\qquad(1)
\displaystyle \text{It is given that (1) is perpendicular to the plane }2x+3y-2z=5.\ \text{So,}
\displaystyle 2a+3b-2c=0.\qquad(2)
\displaystyle \text{It is given that (1) is perpendicular to the plane }x+2y-3z=8.\ \text{So,}
\displaystyle a+2b-3c=0.\qquad(3)
\displaystyle \text{Solving (1), (2) and (3), we get}
\displaystyle \begin{vmatrix}x-1&y+1&z-2\\2&3&-2\\1&2&-3\end{vmatrix}=0.
\displaystyle \Rightarrow -5(x-1)+4(y+1)+1(z-2)=0.
\displaystyle \Rightarrow 5x-4y-z=7.

\displaystyle \textbf{Question 13: }~\text{Find the equation of the plane passing through }(a,b,c) \\ \text{ and parallel to the plane }\overrightarrow{r}\cdot(\widehat{i}+\widehat{j}+\widehat{k})=2. \ \hspace{3.0cm}  \text{[CBSE\ 2014 ]}
\displaystyle \text{Answer:}
\displaystyle \text{Substituting } \overrightarrow{r}=x\widehat{i}+y\widehat{j}+z\widehat{k} \text{ in the given equation of the plane, we get}
\displaystyle (x\widehat{i}+y\widehat{j}+z\widehat{k})\cdot(\widehat{i}+\widehat{j}+\widehat{k})=2.
\displaystyle \Rightarrow x+y+z-2=0.\qquad(1)
\displaystyle \text{The equation of a plane which is parallel to plane (1) is of the form}
\displaystyle x+y+z=k.\qquad(2)
\displaystyle \text{It is given that plane (2) is passing through the point }(a,b,c).\ \text{So,}
\displaystyle a+b+c=k.
\displaystyle \text{Substituting this value of } k \text{ in (2), we get}
\displaystyle x+y+z=a+b+c,\ \text{which is the required equation of the plane.}

\displaystyle \textbf{Question 14: }~\text{Find the equation of the plane passing through the point } \\ (-1,3,2)\text{ and perpendicular to each of the planes }x+2y+3z=5\text{ and } \\ 3x+3y+z=0. 
\displaystyle \text{Answer:}
\displaystyle \text{The equation of any plane passing through point }(-1,3,2)\text{ is}
\displaystyle a(x+1)+b(y-3)+c(z-2)=0.\qquad(1)
\displaystyle \text{It is given that (1) is perpendicular to the plane }x+2y+3z=5.\ \text{So,}
\displaystyle a+2b+3c=0.\qquad(2)
\displaystyle \text{It is given that (1) is perpendicular to the plane }3x+3y+z=0.\ \text{So,}
\displaystyle 3a+3b+c=0.\qquad(3)
\displaystyle \text{Solving (1), (2) and (3), we get}
\displaystyle \begin{vmatrix}x+1&y-3&z-2\\1&2&3\\3&3&1\end{vmatrix}=0.
\displaystyle \Rightarrow -7(x+1)+8(y-3)-3(z-2)=0.
\displaystyle \Rightarrow 7x-8y+3z+25=0.

\displaystyle \textbf{Question 15: }~\text{Find the vector equation of the plane through the points } \\ (2,1,-1)\text{ and }(-1,3,4)\text{ and perpendicular to the plane }x-2y+4z=10.\ \hspace{0.5cm}  \text{[CBSE\ 2013]}
\displaystyle \text{Answer:}
\displaystyle \text{The equation of any plane passing through }(2,1,-1)\text{ is}
\displaystyle a(x-2)+b(y-1)+c(z+1)=0.\qquad(1)
\displaystyle \text{It is given that (1) is passing through }(-1,3,4).\ \text{So,}
\displaystyle a(-1-2)+b(3-1)+c(4+1)=0.
\displaystyle \Rightarrow -3a+2b+5c=0.\qquad(2)
\displaystyle \text{It is given that (1) is perpendicular to the plane }x-2y+4z=10.\ \text{So,}
\displaystyle a-2b+4c=0.\qquad(3)
\displaystyle \text{Solving (1), (2) and (3), we get}
\displaystyle \begin{vmatrix}x-2&y-1&z+1\\-3&2&5\\1&-2&4\end{vmatrix}=0.
\displaystyle \Rightarrow 18(x-2)+17(y-1)+4(z+1)=0.
\displaystyle \Rightarrow 18x+17y+4z-49=0.

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