Question

Let two planes $p_1:2x-y+z=2$, and $p_2:x+2y-z=3$ are given. The equation of the plane through the intersection of $P_1$ and $P_2$ and the point $(3,2,1)$ is

• A. $3x-y+2_Z-9=0$
• B. $x-3y+2z+1=0$
• C. $2x-3y+z-1=0$
• D. $4x-3y+2_Z-8=0$

Answer 1

The correct option is B $x-3y+2z+1=0$

Any plane through the intersection of the planes is given by,
$2x-y+z-2+k(x+2y-z-3)=0$
Since, this plane passes through $(3,2,1)$
$6-2+1-2+k(3+4-1-3)=0$, $k=-1.$
Hence, the plane is $x-3y+2z+1=0$.