Question:
Let the plane passing through the point $(-1,0,-2)$ and perpendicular to each of the planes $2 x+y-z=2$ and $x-y-z=3$ be $a x+b y+c z+8=0$. Then the value of $a+b+c$ is equal to:
Correct Option: , 4
Solution:
Normal of req. plane $(2 \hat{\mathrm{i}}+\hat{\mathrm{j}}-\hat{\mathrm{k}}) \times(\hat{\mathrm{i}}-\hat{\mathrm{j}}-\hat{\mathrm{k}})$
$=-2 \hat{i}+\hat{j}-3 \hat{k}$
Equation of plane
$-2(x+1)+1(y-0)-3(z+2)=0$
$-2 x+y-3 z-8=0$
$2 x-y+3 z+8=0$
$a+b+c=4$
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