Question

Let the plane passing through the point $(-1,0,-2)$ and perpendicular to each of the planes $2x+y-z=2$ and $x-y-z=3$ be $ax+by+cz+8=0$. Then the value of $a+b+c$ is equal to

• A. $3$
• B. $5$
• C. $8$
• D. $4$

Answer 1

The correct option is D $4$

Given equation of plane is $ax+by+cz+8=0$
Since it passes through the point$(-1,0,-2)$
$\therefore -a-2c+8=0$
$\Rightarrow a+2c-8=0\cdots \left(1\right)$
and $ax+by+cz+8=0$ is perpendicular to $2x+y-z=2$ and $x-y-z=3$
$\therefore 2a+b-c=0\cdots \left(2\right)$
$a-b-c=0\cdots \left(3\right)$
From $\left(1\right),\left(2\right)$ and $\left(3\right),$ we get
$a=2,b=-1$ and $c=3$
$\therefore a+b+c=4$