The distance from the point (2, -1, -2) to the plane 6x + 2y - 3z + a = 0 is 2 units. Find a.
A) 2
B) -2 or -30
C) -14
D) -10 or -22
E) -14, -18

Question

Quizplus · Accepted Answer

The distance $d$ from a point $(x_0, y_0, z_0)$ to a plane $Ax + By + Cz + D = 0$ is given by the formula $$d = \frac{|Ax_0 + By_0 + Cz_0 + D|}{\sqrt{A^2 + B^2 + C^2}}.$$ Substituting the given point $(2, -1, -2)$ and the plane equation $6x + 2y - 3z + a = 0$, we get $$2 = \frac{|6(2) + 2(-1) - 3(-2) + a|}{\sqrt{6^2 + 2^2 + (-3)^2}}.$$ Simplifying, we find $$2 = \frac{|12 - 2 + 6 + a|}{7}.$$ Thus, $|16 + a| = 14$, leading to two cases: $16 + a = 14$ or $16 + a = -14$. Solving these gives $a = -2$ or $a = -30$, respectively.

The Distance from the Point (2, -1, -2) to the Plane