Question

Equation of the plane passing through the point $(1,1,1)$ and perpendicular to each of the planes $x+2y+3z=7$ and $2x-3y+4z=0$, is

• A. $17x-2y+7z=12$
• B. $17x+2y-7z=12$
• C. $17x+2y+7z=12$
• D. $17x-2y-7z=12$

Answer 1

The correct option is B $17x+2y-7z=12$

Let $ax+by+cz=1$ be the desired plane.
Since, it is perpendicular to $x+2y+3z=7$ & $2x-3y+4z=0$
Therefore, $a+2b+3c=0$ .... (1)
and $2a-3b+4c=0$ ...(2)
and it passes through $(1,1,1)$
Therefore, $a+b+c=1$ ...(3)
Solving $\left(1\right),\left(2\right)$ and $\left(3\right)$ simultaneously, we get

$a=\frac{17}{12}$, $b=\frac{1}{6}$, $c=-\frac{7}{12}$
Therefore, desired plane is $17x+2y-7z=12$
Ans: B