Let the image of the point P(2, -1, 3) in the plane $x + 2y - z = 0$ be Q. Then the distance of the plane $3x + 2y + z + 29 = 0$ from the point Q is
Step-by-Step Solution
Key Concept: Find image Q of P in plane, then compute distance from Q to second plane.
Foot of perpendicular: $t=-\frac{2-2-3}{1+4+1}=\frac{3}{6}=\frac{1}{2}$. Foot $=\left(\frac{5}{2},0,\frac{5}{2}\right)$. $Q = (3,1,2)$. Distance from $3(3)+2(1)+2+29 = 9+2+2+29=42$. $d=\frac{42}{\sqrt{14}} = 3\sqrt{14}$. Answer: (4)
Correct Answer: $3\sqrt{14}$