Find the coordinates of the point where the line through the points (3, -8, -9) and (2, -7, 1) crosses the plane determined by the points (1, 2, 3), (4, 2, -7), and (0, 4, 3).

Given that the plane passes through the points A(1, 2, 3), B(4, 2, -7), and C(0, 4, 3).
BA = 3i - 9k, CA = -i + 2j
The vector n = BA × CA = 18i + 6j + 6k will be perpendicular to the plane passing through points A, B, C.
So the directional ratios of the plane will be 12, 6, 6.
The equation of the plane will be 12x + 6y + 6z = d, and it passes through point A(1, 2, 3).
So the equation of the plane will be →r ⋅ n = →a ⋅ n, i.e., 2x + y + z = 7.
The equation of the line passing through points D(3, -8, -9) and E(2, -7, 1) is
x - 3 = (y + 8)/(-1) = (z + 9)/(-10)
So the point on the line will be in the form of (t + 3, -8 - t, -9 - 10t), and this point will lie on the plane.
So we have 2(t + 3) + (-8 - t) + (-9 - 10t) = 7
→ 2t + 6 - 8 - t - 9 - 10t = 7
→ -9t - 11 = 7
→ -9t = 18
→ t = -2
So the point where the line crosses the plane will be (1, -6, 7).