3D Geometry
Plane through intersection of planes; perpendicular condition
nta_pyq_2023_jan
Grade 12
Question:
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:
$\frac{250}{83}$
$\frac{15}{53}$
$\frac{25}{83}$
$\frac{250}{82}$
Step-by-Step Solution
Key Concept: Write $P = P_1 + kP_2$, impose perpendicularity with $2x+y-z+1=0$, find $k$, get plane equation, compute distance.
After finding P and computing distance: $d^2 = 250/83$. Answer: (1)
Correct Answer: $\frac{250}{83}$