Determine vector equation of a plane

with the point A(1, 1, 2) and perpendicular to the line joining B(2, 1, -6) & C(-2, 1, 5).

I can't figure out how to do this in 3 dimensions. I'm given one point on the plane. The line gives me 2 points (not necessarily on the plane) and one vector right? How do I find the vector perpendicular to that one vector? and where do I get a second vector from?

A plane is usually determined in either of two ways. In both cases, you need a point on the plane, which you're given. The extra information you need is either (1) the normal vector or (2) two vectors lying in the plane that aren't multiples of each other. You seem to be confusing these two determinations. You're given the normal vector as (-2 - 2, 1 - 1, 5 - (-6)) = (-4, 0, 11). The equation is then just

(-4, 0, 11) dot (x, y, z) = d

for a constant d. Since (1, 1, 2) satisfies this equation, we can find d:
(-4, 0, 11) dot (1, 1, 2) = -4 + 22 = 18 = d.

If all points were on plane as stated in a previous answer, then there's no way the plane could be perpendicular to the line joining B and C. Here you are given not 3 points, or 1 point and 2 vectors parallel to plane, but 1 point and normal vector (C - B or B - C).

Recall that equation of plane passing through point (x0, y0, z0) and with normal vector (a, b, c) has equation: a(x-x0) + b(y-y0) + c(z-z0) = 0

Point on plane: A(1, 1, 2)
Normal vector: (2, 1, -6) - (-2, 1, 5) = (4, 0, -11)

Equation of plane:
4(x-1) + 0(y-1) - 11(z-2) = 0
4x - 4 + 0 - 11z + 22 = 0
4x - 11z = -18

EDIT: Sorry, I just realized you need vector equation of plane:
(4, 0, -11) • (x, y, z) = (4, 0, -11) • (1, 1, 2)
(4, 0, -11) • (x, y, z) = -18

I assume all the points are on the plane, else the question makes no sense!

one vector AB = B - A =[1, 0, -8] the other AC = C -A =[-3, 0, 3]