Question

The equation of the plane passing through the point (1, 2, -3) and perpendicular to the planes $3 x+y-2 z=5$ and $2 x-5 y-z=7$ is

• A. $3 x-10 y-2 z+11=0$
• B. $6 x-5 y-2 z-2=0$
• C. $11 x+y+17 z+38=0$
• D. $6 x-5 y+2 z+10=0$

Answer 1

Answer: (C)

Normal vector :
$\begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\ 3&1&-2\\ 2&-5&-1\end{vmatrix}$
$ = -11 \hat{i} - \hat{j} + 17 \hat{k}$
So drs of normal to the required plane is
$<11,1,17>$
plane passes through (1,2,-3) So eq $^{ n }$ of plane:
$ 11(x-1)+1(y-2)+17(z+3)=0 $
$\Rightarrow 11 x+y+17 z+38=0$