Question:
The plane containing the line $\frac{x-3}{2}=\frac{y+2}{-1}=\frac{z-1}{3}$ and also containing its projection on the plane
$2 x+3 y-z=5$, contains which one of the following points ?
Correct Option: 1
Solution:
The normal vector of required plane
$=(2 \hat{\mathrm{i}}-\hat{\mathrm{j}}+3 \hat{\mathrm{k}}) \times(2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}-\hat{\mathrm{k}})$
$=-8 \hat{\mathrm{i}}+8 \hat{\mathrm{j}}+8 \hat{\mathrm{k}}$
So, direction ratio of normal is $(-1,1,1)$ So required plane is
$-(x-3)+(y+2)+(z-1)=0$
$\Rightarrow-x+y+z+4=0$
Which is satisfied by (2, 0, –2)