Posted by Grace on Saturday, January 7, 2012 at 9:34pm.
At the point were a line intersects a plane [with the equation (24x+32y+40z=480) and three points of A(5, 10, 1), B(6, 3, 6), and C(12, 1, 4)], the vector i, vector j, and vector kcoefficients of the line equal the corresponding values of x, y, and z for the plane. By substituting these coefficients, for x, y, and z in the equation for the plane, find the directed distance, d, from the known point (7, 11, 3) to the intersection point (x, y, z).

Precalc  incomplete  Steve, Sunday, January 8, 2012 at 3:42pm
You have two vectors:
a = (7,11,3)
b = (x,y,z)
The vector from a to b can be found by
a + d = b
so, d = b  a
= (x7,y11,z3)
so the distance is d in the direction of d.