\displaystyle \textbf{Question 1: }~\text{Find the equation of the plane which is parallel to } \\ 2x-3y+z=0 \text{ and which passes through }(1,-1,2).
\displaystyle \text{Answer:}
\displaystyle \text{Let the equation of a plane parallel to the given plane be}
\displaystyle 2x-3y+z=k.\qquad(1)
\displaystyle \text{This passes through }(1,-1,2).\ \text{So,}
\displaystyle 2(1)-3(-1)+2=k.
\displaystyle \Rightarrow k=7.
\displaystyle \text{Substituting this in (1), we get}
\displaystyle 2x-3y+z=7,\ \text{which is the equation of the required plane.}

\displaystyle \textbf{Question 2: }~\text{Find the equation of the plane through }(3,4,-1)\text{ which is parallel} \\ \text{to the plane }\overrightarrow{r}\cdot(2\widehat{i}-3\widehat{j}+5\widehat{k})+2=0.
\displaystyle \text{Answer:}
\displaystyle \text{Let the equation of a plane parallel to the given plane be}
\displaystyle \overrightarrow{r}\cdot(2\widehat{i}-3\widehat{j}+5\widehat{k})=k.\qquad(1)
\displaystyle (x\widehat{i}+y\widehat{j}+z\widehat{k})\cdot(2\widehat{i}-3\widehat{j}+5\widehat{k})=k.
\displaystyle \text{This passes through }(3,4,-1).\ \text{So,}
\displaystyle (3\widehat{i}+4\widehat{j}-\widehat{k})\cdot(2\widehat{i}-3\widehat{j}+5\widehat{k})=k.
\displaystyle \Rightarrow k=6-12-5=-11.
\displaystyle \text{Substituting this in (1), we get}
\displaystyle \overrightarrow{r}\cdot(2\widehat{i}-3\widehat{j}+5\widehat{k})=-11.
\displaystyle \Rightarrow \overrightarrow{r}\cdot(2\widehat{i}-3\widehat{j}+5\widehat{k})+11=0,\ \text{which is the equation of the required plane.}

\displaystyle \textbf{Question 3: }~\text{Find the equation of the plane passing through the line of} \\ \text{intersection of the planes }2x-7y+4z-3=0,\ 3x-5y+4z+11=0\text{ and the point }(-2,1,3).
\displaystyle \text{Answer:}
\displaystyle \text{The equation of the plane passing through the line of intersection of the given planes is}
\displaystyle 2x-7y+4z-3+\lambda(3x-5y+4z+11)=0.\qquad(1)
\displaystyle \text{This passes through }(-2,1,3).\ \text{So,}
\displaystyle -4-7+12-3+\lambda(-6-5+12+11)=0.
\displaystyle \Rightarrow -2+12\lambda=0.
\displaystyle \Rightarrow \lambda=\frac{1}{6}.
\displaystyle \text{Substituting this in (1), we get}
\displaystyle 2x-7y+4z-3+\frac{1}{6}(3x-5y+4z+11)=0.
\displaystyle \Rightarrow 12x-42y+24z-18+3x-5y+4z+11=0.
\displaystyle \Rightarrow 15x-47y+28z=7.

\displaystyle \textbf{Question 4: }~\text{Find the equation of the plane through the point }\widehat{i}+\widehat{j}+\widehat{k}\text{ and passing} \\ \text{through the line of intersection of the planes }\overrightarrow{r}\cdot(\widehat{i}+3\widehat{j}-\widehat{k})=0\text{ and }\overrightarrow{r}\cdot(\widehat{j}+2\widehat{k})=0.
\displaystyle \text{Answer:}
\displaystyle \text{The equation of the plane passing through the line of intersection of the given planes is}
\displaystyle \overrightarrow{r}\cdot(\widehat{i}+3\widehat{j}-\widehat{k})+\lambda\big(\overrightarrow{r}\cdot(\widehat{j}+2\widehat{k})\big)=0.
\displaystyle \Rightarrow \overrightarrow{r}\cdot\big[\widehat{i}+(3+\lambda)\widehat{j}+(-1+2\lambda)\widehat{k}\big]=0.\qquad(1)
\displaystyle \text{This passes through }2\widehat{i}+\widehat{j}-\widehat{k}.\ \text{So,}
\displaystyle (2\widehat{i}+\widehat{j}-\widehat{k})\cdot\big[\widehat{i}+(3+\lambda)\widehat{j}+(-1+2\lambda)\widehat{k}\big]=0.
\displaystyle \Rightarrow 2+3+\lambda+1-2\lambda=0.
\displaystyle \Rightarrow \lambda=6.
\displaystyle \text{Substituting this in (1), we get}
\displaystyle \overrightarrow{r}\cdot\big[\widehat{i}+(3+6)\widehat{j}+(-1+12)\widehat{k}\big]=0.
\displaystyle \Rightarrow \overrightarrow{r}\cdot(\widehat{i}+9\widehat{j}+11\widehat{k})=0.

\displaystyle \textbf{Question 5: }~\text{Find the equation of the plane passing through the line of intersection} \\ \text{of the planes }2x-y=0\text{ and }3x-y=0\text{ and perpendicular to the plane }4x+5y-3z=8.
\displaystyle \text{Answer:}
\displaystyle \text{The equation of the plane passing through the line of intersection of the given planes is}
\displaystyle 2x-y+\lambda(3z-y)=0.
\displaystyle \Rightarrow 2x+(-1-\lambda)y+3\lambda z=0.\qquad(1)
\displaystyle \text{This plane is perpendicular to }4x+5y-3z=8.\ \text{So,}
\displaystyle 2(4)+(-1-\lambda)5-9\lambda=0.
\displaystyle \Rightarrow 8-5-5\lambda-9\lambda=0.
\displaystyle \Rightarrow \lambda=\frac{3}{14}.
\displaystyle \text{Substituting this in (1), we get}
\displaystyle 2x+\left(-1-\frac{3}{14}\right)y+3\left(\frac{3}{14}\right)z=0.
\displaystyle \Rightarrow 28x-17y+9z=0.

\displaystyle \textbf{Question 6: }~\text{Find the equation of the plane which contains the line of intersection} \\ \text{of the planes }x+2y+3z-4=0\text{ and }2x+y-z+5=0\text{ and which is perpendicular} \\ \text{to the plane }5x+3y-6z+8=0.
\displaystyle \text{Answer:}
\displaystyle \text{The equation of the plane passing through the line of intersection of the given planes is}
\displaystyle x+2y+3z-4+\lambda(2x+y-z+5)=0.
\displaystyle \Rightarrow (1+2\lambda)x+(2+\lambda)y+(3-\lambda)z-4+5\lambda=0.\qquad(1)
\displaystyle \text{This plane is perpendicular to }5x+3y-6z+8=0.\ \text{So,}
\displaystyle 5(1+2\lambda)+3(2+\lambda)-6(3-\lambda)=0.
\displaystyle \Rightarrow 5+10\lambda+6+3\lambda-18+6\lambda=0.
\displaystyle \Rightarrow 19\lambda-7=0.
\displaystyle \Rightarrow \lambda=\frac{7}{19}.
\displaystyle \text{Substituting this in (1), we get}
\displaystyle \left(1+2\frac{7}{19}\right)x+\left(2+\frac{7}{19}\right)y+\left(3-\frac{7}{19}\right)z-4+5\frac{7}{19}=0.
\displaystyle \Rightarrow 33x+45y+50z-41=0.

\displaystyle \textbf{Question 7: }~\text{Find the equation of the plane through the line of intersection} \\ \text{of the planes }x+2y+3z-4=0\text{ and }x-y+z+3=0\text{ and passing through the origin.}
\displaystyle \text{Answer:}
\displaystyle \text{The equation of the plane passing through the line of intersection of the given planes is}
\displaystyle x+2y+3z+4+\lambda(x-y+z+3)=0.\qquad(1)
\displaystyle \text{This passes through }(0,0,0).\ \text{So,}
\displaystyle 0+0+0+4+\lambda(0-0+0+3)=0.
\displaystyle \Rightarrow 4+3\lambda=0.
\displaystyle \Rightarrow \lambda=-\frac{4}{3}.
\displaystyle \text{Substituting this in (1), we get}
\displaystyle x+2y+3z+4-\frac{4}{3}(x-y+z+3)=0.
\displaystyle \Rightarrow -x+10y+5z=0.
\displaystyle \Rightarrow x-10y-5z=0.

\displaystyle \textbf{Question 8: }~\text{Find the vector equation in scalar product form of the plane containing} \\ \text{the line of intersection of the planes }x-3y+2z-5=0\text{ and }2x-y+3z-1=0\text{ and passing through }(1,-2,3).
\displaystyle \text{Answer:}
\displaystyle \text{The equation of the plane passing through the line of intersection of the given planes is}
\displaystyle x-3y+2z-5+\lambda(2x-y+3z-1)=0.\qquad(1)
\displaystyle \text{This passes through }(1,-2,3).\ \text{So,}
\displaystyle 1+6+6-5+\lambda(2+2+9-1)=0.
\displaystyle \Rightarrow 8+12\lambda=0.
\displaystyle \Rightarrow \lambda=-\frac{2}{3}.
\displaystyle \text{Substituting this in (1), we get}
\displaystyle x-3y+2z-5-\frac{2}{3}(2x-y+3z-1)=0.
\displaystyle \Rightarrow -x-7y-13=0.
\displaystyle \Rightarrow x+7y+13=0.
\displaystyle \Rightarrow \overrightarrow{r}\cdot(\widehat{i}+7\widehat{j})+13=0,\ \text{which is the required vector equation of the plane.}

\displaystyle \textbf{Question 9: }~\text{Find the equation of the plane which is perpendicular to the plane } \\ 5x+3y+6z+8=0\text{ and which contains the line of intersection of the planes }x+2y+3z-4=0,\ 2x+y-z+5=0.
\displaystyle \text{Answer:}
\displaystyle \text{The equation of the plane passing through the line of intersection of the given planes is}
\displaystyle x+2y+3z-4+\lambda(2x+y-z+5)=0.
\displaystyle \Rightarrow (1+2\lambda)x+(2+\lambda)y+(3-\lambda)z-4+5\lambda=0.\qquad(1)
\displaystyle \text{This plane is perpendicular to }5x+3y+6z+8=0.\ \text{So,}
\displaystyle 5(1+2\lambda)+3(2+\lambda)+6(3-\lambda)=0.
\displaystyle \Rightarrow 5+10\lambda+6+3\lambda+18-6\lambda=0.
\displaystyle \Rightarrow 7\lambda+29=0.
\displaystyle \Rightarrow \lambda=-\frac{29}{7}.
\displaystyle \text{Substituting this in (1), we get}
\displaystyle \left(1+2\left(-\frac{29}{7}\right)\right)x+\left(2-\frac{29}{7}\right)y+\left(3+\frac{29}{7}\right)z-4+5\left(-\frac{29}{7}\right)=0.
\displaystyle \Rightarrow -51x-15y+50z-173=0.
\displaystyle \Rightarrow 51x+15y-50z+173=0.

\displaystyle \textbf{Question 10: }~\text{Find the equation of the plane through the line of intersection} \\ \text{of the planes }\overrightarrow{r}\cdot(\widehat{i}+3\widehat{j})+6=0\text{ and }\overrightarrow{r}\cdot(3\widehat{i}-\widehat{j}-4\widehat{k})=0,\text{ which is at} \\ \text{a unit distance from the origin. \ [CBSE\ 2010,\ 2013]}
\displaystyle \text{Answer:}
\displaystyle \text{The equation of the plane passing through the line of intersection of the given planes is}
\displaystyle \overrightarrow{r}\cdot(\widehat{i}+3\widehat{j})+6+\lambda\big(\overrightarrow{r}\cdot(3\widehat{i}-\widehat{j}-4\widehat{k})\big)=0.
\displaystyle \Rightarrow \overrightarrow{r}\cdot\big[(1+3\lambda)\widehat{i}+(3-\lambda)\widehat{j}-4\lambda\widehat{k}\big]+6=0.\qquad(1)
\displaystyle \Rightarrow \overrightarrow{r}\cdot\big[(1+3\lambda)\widehat{i}+(3-\lambda)\widehat{j}-4\lambda\widehat{k}\big]=-6.
\displaystyle \Rightarrow \overrightarrow{r}\cdot\big[(-1-3\lambda)\widehat{i}+(\lambda-3)\widehat{j}+4\lambda\widehat{k}\big]=6.
\displaystyle \text{Dividing both sides by }\sqrt{(-1-3\lambda)^2+(\lambda-3)^2+16\lambda^2}\text{, we get}
\displaystyle \overrightarrow{r}\cdot\left(\frac{(-1-3\lambda)\widehat{i}+(\lambda-3)\widehat{j}+4\lambda\widehat{k}}{\sqrt{(-1-3\lambda)^2+(\lambda-3)^2+16\lambda^2}}\right)=\frac{6}{\sqrt{(-1-3\lambda)^2+(\lambda-3)^2+16\lambda^2}}.
\displaystyle \text{The perpendicular distance of plane (1) from the origin is }\frac{6}{\sqrt{(-1-3\lambda)^2+(\lambda-3)^2+16\lambda^2}}.
\displaystyle \Rightarrow 1=\frac{6}{\sqrt{(-1-3\lambda)^2+(\lambda-3)^2+16\lambda^2}}\ (\text{Given}).
\displaystyle \Rightarrow \sqrt{(-1-3\lambda)^2+(\lambda-3)^2+16\lambda^2}=6.
\displaystyle \Rightarrow 1+9\lambda^2+6\lambda+\lambda^2-6\lambda+9+16\lambda^2=36.
\displaystyle \Rightarrow 26\lambda^2-26=0.
\displaystyle \Rightarrow \lambda^2=1.
\displaystyle \Rightarrow \lambda=1,-1.
\displaystyle \text{Case 1: Substituting }\lambda=1\text{ in (1), we get}
\displaystyle \overrightarrow{r}\cdot(4\widehat{i}+2\widehat{j}-4\widehat{k})+6=0.
\displaystyle \text{Case 2: Substituting }\lambda=-1\text{ in (1), we get}
\displaystyle \overrightarrow{r}\cdot(-2\widehat{i}+4\widehat{j}+4\widehat{k})+6=0.

\displaystyle \textbf{Question 11: }~\text{Find the equation of the plane passing through the intersection of} \\ \text{the planes }2x+3y-z+1=0\text{ and }x+y-2z+3=0\text{ and perpendicular to the plane }3x-y-2z-4=0.
\displaystyle \text{Answer:}
\displaystyle \text{The equation of the plane passing through the line of intersection of the given planes is}
\displaystyle 2x+3y-z+1+\lambda(x+y-2z+3)=0.
\displaystyle \Rightarrow (2+\lambda)x+(3+\lambda)y+(-1-2\lambda)z+1+3\lambda=0.\qquad(1)
\displaystyle \text{This plane is perpendicular to }3x-y-2z-4=0.
\displaystyle \Rightarrow 3(2+\lambda)-1(3+\lambda)-2(-1-2\lambda)=0.
\displaystyle \Rightarrow 6+3\lambda-3-\lambda+2+4\lambda=0.
\displaystyle \Rightarrow 6\lambda+5=0.
\displaystyle \Rightarrow \lambda=-\frac{5}{6}.
\displaystyle \text{Substituting this in (1), we get}
\displaystyle \left(2-\frac{5}{6}\right)x+\left(3-\frac{5}{6}\right)y+\left(-1-2\left(-\frac{5}{6}\right)\right)z+1+3\left(-\frac{5}{6}\right)=0.
\displaystyle \Rightarrow \frac{7}{6}x+\frac{13}{6}y+\frac{4}{6}z-\frac{9}{6}=0.
\displaystyle \Rightarrow 7x+13y+4z-9=0.

\displaystyle \textbf{Question 12: }~\text{Find the equation of the plane that contains the line of intersection} \\ \text{of the planes }\overrightarrow{r}\cdot(\widehat{i}+2\widehat{j}+3\widehat{k})-4=0\text{ and }\overrightarrow{r}\cdot(2\widehat{i}+\widehat{j}-\widehat{k})+5=0\text{ and which is perpendicular} \\ \text{to the plane }\overrightarrow{r}\cdot(5\widehat{i}+3\widehat{j}-6\widehat{k})+8=0.\ \text{[CBSE\ 2011,\ 2013]}
\displaystyle \text{Answer:}
\displaystyle \text{The equation of the plane passing through the line of intersection of the given planes is}
\displaystyle \vec r\cdot(\hat i+2\hat j+3\hat k)-4+\lambda[\vec r\cdot(2\hat i+\hat j-\hat k)+5]=0.
\displaystyle \vec r\cdot[(1+2\lambda)\hat i+(2+\lambda)\hat j+(3-\lambda)\hat k]-4+5\lambda=0.\qquad(1)
\displaystyle \text{This plane is perpendicular to }\vec r\cdot(5\hat i+3\hat j-6\hat k)+8=0.
\displaystyle \Rightarrow 5(1+2\lambda)+3(2+\lambda)-6(3-\lambda)=0.
\displaystyle \Rightarrow 5+10\lambda+6+3\lambda-18+6\lambda=0.
\displaystyle \Rightarrow 19\lambda-7=0.
\displaystyle \Rightarrow \lambda=\frac{7}{19}.
\displaystyle \text{Substituting this in (1), we get}
\displaystyle \vec r\cdot\left[\left(1+2\frac{7}{19}\right)\hat i+\left(2+\frac{7}{19}\right)\hat j+\left(3-\frac{7}{19}\right)\hat k\right]-4+5\left(\frac{7}{19}\right)=0.
\displaystyle \Rightarrow \vec r\cdot(33\hat i+45\hat j+50\hat k)-41=0.
\displaystyle \Rightarrow (x\hat i+y\hat j+z\hat k)\cdot(33\hat i+45\hat j+50\hat k)-41=0.
\displaystyle \Rightarrow 33x+45y+50z-41=0.

\displaystyle \textbf{Question 13: }~\text{Find the vector equation of the plane passing through the intersection} \\ \text{of the planes }\overrightarrow{r}\cdot(\widehat{i}+\widehat{j}+\widehat{k})=6\text{ and }\overrightarrow{r}\cdot(2\widehat{i}+3\widehat{j}+4\widehat{k})=-5\text{ and the point }(1,1,1).\ \text{[CBSE\ 2014]}
\displaystyle \text{Answer:}
\displaystyle \text{The equation of the plane passing through the line of intersection of the given planes is}
\displaystyle \vec r\cdot(\hat i+\hat j+\hat k)-6+\lambda\left(\vec r\cdot(2\hat i+3\hat j+4\hat k)+5\right)=0.
\displaystyle \vec r\cdot\left[(1+2\lambda)\hat i+(1+3\lambda)\hat j+(1+4\lambda)\hat k\right]-6+5\lambda=0.\qquad(1)
\displaystyle \text{This passes through }\hat i+\hat j+\hat k.\ \text{So,}
\displaystyle (\hat i+\hat j+\hat k)\cdot\left[(1+2\lambda)\hat i+(1+3\lambda)\hat j+(1+4\lambda)\hat k\right]-6+5\lambda=0.
\displaystyle \Rightarrow 1+2\lambda+1+3\lambda+1+4\lambda-6+5\lambda=0.
\displaystyle \Rightarrow 14\lambda-3=0.
\displaystyle \Rightarrow \lambda=\frac{3}{14}.
\displaystyle \text{Substituting this in (1), we get}
\displaystyle \vec r\cdot\left[\left(1+2\left(\frac{3}{14}\right)\right)\hat i+\left(1+3\left(\frac{3}{14}\right)\right)\hat j+\left(1+4\left(\frac{3}{14}\right)\right)\hat k\right]-6+5\left(\frac{3}{14}\right)=0.
\displaystyle \Rightarrow \vec r\cdot\left(\frac{10}{7}\hat i+\frac{23}{14}\hat j+\frac{13}{7}\hat k\right)-\frac{69}{14}=0.
\displaystyle \Rightarrow \vec r\cdot(20\hat i+23\hat j+26\hat k)=69.

\displaystyle \textbf{Question 14: }~\text{Find the equation of the plane passing through the intersection} \\ \text{of the planes }\overrightarrow{r}\cdot(2\widehat{i}+\widehat{j}+3\widehat{k})=7,\ \overrightarrow{r}\cdot(2\widehat{i}+5\widehat{j}+3\widehat{k})=9\text{ and the point }(2,1,3).\ \text{[CBSE\ 2007]}
\displaystyle \text{Answer:}
\displaystyle \text{The equation of the plane passing through the line of intersection of the given planes is}
\displaystyle \vec r\cdot(2\hat i+\hat j+3\hat k)-7+\lambda\left[\vec r\cdot(2\hat i+5\hat j+3\hat k)-9\right]=0.
\displaystyle \vec r\cdot\left[(2+2\lambda)\hat i+(1+5\lambda)\hat j+(3+3\lambda)\hat k\right]-7-9\lambda=0.\qquad(1)
\displaystyle \text{This passes through }(2,1,3).\text{ So,}
\displaystyle (2\hat i+\hat j+3\hat k)\cdot\left[(2+2\lambda)\hat i+(1+5\lambda)\hat j+(3+3\lambda)\hat k\right]-7-9\lambda=0.
\displaystyle \Rightarrow 4+4\lambda+1+5\lambda+9+9\lambda-7-9\lambda=0.
\displaystyle \Rightarrow 9\lambda+7=0.
\displaystyle \Rightarrow \lambda=-\frac{7}{9}.
\displaystyle \text{Substituting this in (1), we get}
\displaystyle \vec r\cdot\left[\left(2+2\left(-\frac79\right)\right)\hat i+\left(1+5\left(-\frac79\right)\right)\hat j+\left(3+3\left(-\frac79\right)\right)\hat k\right]-7-9\left(-\frac79\right)=0.
\displaystyle \Rightarrow \vec r\cdot\left(\frac49\hat i-\frac{26}{9}\hat j+\frac23\hat k\right)=0.
\displaystyle \Rightarrow \vec r\cdot(4\hat i-26\hat j+6\hat k)=0.
\displaystyle \Rightarrow \vec r\cdot(2\hat i-13\hat j+3\hat k)=0.

\displaystyle \textbf{Question 15: }~\text{Find the equation of the plane through the intersection} \\ \text{of the planes }3x-y+2z=4\text{ and }x+y+z=2\text{ and the point }(2,2,1).\ \text{[NCERT]}
\displaystyle \text{Answer:}
\displaystyle \text{The equation of the plane passing through the line of intersection of the given planes is}
\displaystyle 3x-y+2z-4+\lambda(x+y+z-2)=0.\qquad(1)
\displaystyle \text{This passes through }(2,2,1).\text{ So,}
\displaystyle 6-2+2-4+\lambda(2+2+1-2)=0.
\displaystyle \Rightarrow 2+3\lambda=0.
\displaystyle \Rightarrow \lambda=-\frac23.
\displaystyle \text{Substituting this in (1), we get}
\displaystyle 3x-y+2z-4-\frac23(x+y+z-2)=0.
\displaystyle \Rightarrow 9x-3y+6z-12-2x-2y-2z+4=0.
\displaystyle \Rightarrow 7x-5y+4z=8.

\displaystyle \textbf{Question 16: }~\text{Find the vector equation of the plane through the line of intersection} \\ \text{of the planes }x+y+z=1\text{ and }2x+3y+4z=5\text{ which is perpendicular to the plane }x-y+z=0.\ \text{[CBSE\ 2014]}
\displaystyle \text{Answer:}
\displaystyle P_1:\ x+y+z=1\ \Rightarrow\ x+y+z-1=0
\displaystyle P_2:\ 2x+3y+4z=5\ \Rightarrow\ 2x+3y+4z-5=0
\displaystyle \text{Plane through the line of intersection of }P_1\text{ and }P_2\text{ is }P_1+\lambda P_2=0
\displaystyle (x+y+z-1)+\lambda(2x+3y+4z-5)=0
\displaystyle \text{Normal vector of this plane is }\overrightarrow n=(1,1,1)+\lambda(2,3,4)=(1+2\lambda,\ 1+3\lambda,\ 1+4\lambda)
\displaystyle \text{Given plane: }x-y+z=0\ \Rightarrow\ \text{normal vector }\overrightarrow m=(1,-1,1)
\displaystyle \text{Required plane is perpendicular to }x-y+z=0\ \Rightarrow\ \overrightarrow n\cdot\overrightarrow m=0
\displaystyle (1+2\lambda)\cdot1+(1+3\lambda)\cdot(-1)+(1+4\lambda)\cdot1=0
\displaystyle (1+2\lambda)-(1+3\lambda)+(1+4\lambda)=0
\displaystyle 1+3\lambda=0
\displaystyle \lambda=-\frac{1}{3}
\displaystyle (x+y+z-1)-\frac{1}{3}(2x+3y+4z-5)=0
\displaystyle 3(x+y+z-1)-(2x+3y+4z-5)=0
\displaystyle 3x+3y+3z-3-2x-3y-4z+5=0
\displaystyle x-z+2=0
\displaystyle \text{Vector form: }\vec r\cdot(\hat i-\hat k)=-2

\displaystyle \textbf{Question 17: }~\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.
\displaystyle \text{Answer:}
\displaystyle \vec r\cdot(\hat i+\hat j+\hat k)=2
\displaystyle \vec r=x\hat i+y\hat j+z\hat k
\displaystyle x+y+z=2
\displaystyle \text{Normal vector }=(1,1,1)
\displaystyle \text{Equation of plane parallel to it: }x+y+z=d
\displaystyle \text{Since it passes through }(a,b,c)
\displaystyle a+b+c=d
\displaystyle \therefore\ x+y+z=a+b+c

\displaystyle \textbf{Question 18: }~\text{Find the equation of the plane which contains the line of intersection} \\ \text{of the planes }x+2y+3z-4=0\text{ and }2x+y-z+5=0\text{ and whose } \\ x\text{-intercept is twice its }z\text{-intercept.} \text{Hence, write the equation of the plane} \\ \text{passing through the point }(2,3,-1)\text{ and parallel to} \\ \text{the plane obtained above. \ [CBSE\ 2016]}
\displaystyle \text{Answer:}
\displaystyle \text{The equation of the family of planes passing through the intersection of the planes }x+2y+3z-4=0\text{ and }2x+y-z+5=0\text{ is}
\displaystyle (x+2y+3z-4)+k(2x+y-z+5)=0,
\displaystyle \Rightarrow (2k+1)x+(k+2)y+(3-k)z=4-5k.
\displaystyle \Rightarrow \frac{x}{\frac{4-5k}{2k+1}}+\frac{y}{\frac{4-5k}{k+2}}+\frac{z}{\frac{4-5k}{3-k}}=1.
\displaystyle \text{It is given that the }x\text{-intercept of the required plane is twice its }z\text{-intercept.}
\displaystyle \frac{4-5k}{2k+1}=2\left(\frac{4-5k}{3-k}\right).
\displaystyle \Rightarrow (4-5k)(3-k)=(4k+2)(4-5k).
\displaystyle \Rightarrow (4-5k)(3-k-4k-2)=0.
\displaystyle \Rightarrow (4-5k)(1-5k)=0.
\displaystyle \Rightarrow 4-5k=0\text{ or }1-5k=0.
\displaystyle \Rightarrow k=\frac45\text{ or }k=\frac15.
\displaystyle \text{When }k=\frac45,\text{ the equation of the plane is}
\displaystyle \left(2\cdot\frac45+1\right)x+\left(\frac45+2\right)y+\left(3-\frac45\right)z=4-5\cdot\frac45.
\displaystyle \Rightarrow \frac{13}{5}x+\frac{14}{5}y+\frac{11}{5}z=0.
\displaystyle \Rightarrow 13x+14y+11z=0.
\displaystyle \text{This plane does not satisfy the given condition, so it is rejected.}
\displaystyle \text{When }k=\frac15,\text{ the equation of the plane is}
\displaystyle \left(2\cdot\frac15+1\right)x+\left(\frac15+2\right)y+\left(3-\frac15\right)z=4-5\cdot\frac15.
\displaystyle \Rightarrow \frac75x+\frac{11}{5}y+\frac{14}{5}z=3.
\displaystyle \Rightarrow 7x+11y+14z=15.
\displaystyle \text{Thus, the equation of the required plane is }7x+11y+14z=15.
\displaystyle \text{Also, the equation of the plane passing through }(2,3,-1)\text{ and parallel to }7x+11y+14z=15\text{ is}
\displaystyle 7(x-2)+11(y-3)+14(z+1)=0.
\displaystyle \Rightarrow 7x+11y+14z=33.

\displaystyle \textbf{Question 19: }~\text{Find the equation of the plane through the line of intersection} \\ \text{of the planes }x+y+z=1\text{ and }2x+3y+4z=5\text{ and twice of its }y\text{-intercept is equal to} \\ \text{three times its }z\text{-intercept. \ [CBSE\ 2017]}
\displaystyle \text{Answer:}
\displaystyle \text{The equation of the family of the planes passing through the intersection of the planes }x+y+z=1\text{ and }2x+3y+4z=5\text{ is}
\displaystyle (x+y+z-1)+k(2x+3y+4z-5)=0,
\displaystyle \Rightarrow (2k+1)x+(3k+1)y+(4k+1)z=5k+1.\qquad(1)
\displaystyle \Rightarrow \frac{x}{\frac{5k+1}{2k+1}}+\frac{y}{\frac{5k+1}{3k+1}}+\frac{z}{\frac{5k+1}{4k+1}}=1.
\displaystyle \text{It is given that twice of the }y\text{-intercept is equal to three times its }z\text{-intercept.}
\displaystyle \Rightarrow 2\left(\frac{5k+1}{3k+1}\right)=3\left(\frac{5k+1}{4k+1}\right).
\displaystyle \Rightarrow (5k+1)(8k+2-9k-3)=0.
\displaystyle \Rightarrow (5k+1)(-k-1)=0.
\displaystyle \Rightarrow (5k+1)(k+1)=0.
\displaystyle \Rightarrow 5k+1=0\ \text{or}\ k+1=0.
\displaystyle \Rightarrow k=-\frac{1}{5}\ \text{or}\ k=-1.
\displaystyle \text{Putting }k=-\frac{1}{5}\text{ in (1), we get}
\displaystyle \left(2\left(-\frac{1}{5}\right)+1\right)x+\left(3\left(-\frac{1}{5}\right)+1\right)y+\left(4\left(-\frac{1}{5}\right)+1\right)z=5\left(-\frac{1}{5}\right)+1.
\displaystyle \Rightarrow \frac{3}{5}x+\frac{2}{5}y+\frac{1}{5}z=0.
\displaystyle \Rightarrow 3x+2y+z=0.
\displaystyle \text{This plane passes through the origin, so the intercepts are zero and the condition is not satisfied.}
\displaystyle \text{Putting }k=-1\text{ in (1), we get}
\displaystyle (-2+1)x+(-3+1)y+(-4+1)z=5(-1)+1.
\displaystyle \Rightarrow -x-2y-3z=-4.
\displaystyle \Rightarrow x+2y+3z=4.
\displaystyle \text{Here, twice of the }y\text{-intercept equals three times the }z\text{-intercept.}
\displaystyle \text{Thus, the equation of the required plane is }x+2y+3z=4.

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