Find the coordinates of the point where the line through the points (3, -8, -9) and (2, -7, 1) crosses the plane 3x + 2y + z + 14 = 0.

The line passes through the points (3, -8, -9) and (2, -7, 1).
The equation of the line can be written as
(x - 3)/(-1) = (y + 8)/(-1) = (z + 9)/(-10) = t
Thus, a point on the line can be written as (t + 2, -t - 7, -10t + 1).
This point lies on the plane and so,
3(t + 2) + 2(-t - 7) + (-10t + 1) + 14 = 0
3t + 6 - 2t - 14 - 10t + 1 + 14 = 0
-9t + 7 = 0
t = 7/9
Re-substituting t, we get the point as
(7/9 + 2, -7/9 - 7, -70/9 + 1) = (25/9, -70/9, -61/9)