Let the plane $P$ pass through the intersection of the planes $2x+3y-z=2$ and $x+2y+3z=6$, and be perpendicular to the plane $2x+y-z+1=0$. If $d$ is the distance of $P$ from the point $(-7,1,1)$, then $d^2$ is equal to :

Question

Dr. Abubacker · Accepted Answer

The correct answer is Option A: $\frac{250}{83}$.

Q. Let the plane $P$ pass through the intersection of the planes $2x+3y-z=2$ and $x+2y+3z=6$, and be perpendicular to the plane $2x+y-z+1=0$. If $d$ is the distance of $P$ from the point $(-7,1,1)$, then $d^2$ is equal to :

Explanation

Reviewed by Dr. Abubacker

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