The distance of the point (2, 3, 4) from the plane 3x + 2y + 2z + 5 = 0 measured parallel to the line $\frac{x+3}{3}=\frac{y-2}{6}=\frac{z}{2}$ is

The correct option is C

7

Equation of a line through p(2, 3, 4) and parallel to the given line is

$\frac{x-2}{3}=\frac{y-3}{6}=\frac{z-4}{2}\lambda$ (say)

Let Q $(3\lambda +2,6\lambda +3,2\lambda +4)$ is the point of intersection with the plane.

$\therefore$Q lies in the plane $\Rightarrow \lambda$ = - 1 $\Rightarrow$ Q = (-1, -3, 2)

$\therefore$ PQ = 7