Question

The equations of two planes are $P_1:2x-y+z=2$, and $P_2:x+2y-z=3$ . The equation of the plane which passes through the point $(-1,3,2)$ and is perpendicular to each of the planes $P_1$ and $P_2$ is

• A. $x+3y-5_Z+2=0$
• B. $x+3y+5z-18=0$
• C. $x-3y-5z+20=0$
• D. $x-3y+5z=0$

Answer 1

Answer: (C)

$x-3y-5z+20=0$

The dr's of the normal to the $2$ planes are
$(2,-1,1)$ and $(1,2,-1)$
The normal to the third plane will be perpendicular to both these.
$n=AB\times BC$
$n=(1,-3,-5)$
The equation of the plane will be of the form,
$x-3y-5z=d$
Since, the plane passes through $(-1,3,2)$
$d=-20$