If the line drawn from the point $(-2, -1, -3)$ meets a plane at a right angle at the point $(1, -3, 3)$,find the equation of the plane.

(3X - 2Y + 6Z - 27 = 0) The line passes through $P(-2, -1, -3)$ and meets the plane at $Q(1, -3, 3)$ at a right angle.
Thus,the vector $\vec{PQ}$ is normal to the plane.
$\vec{PQ} = (1 - (-2))\hat{i} + (-3 - (-1))\hat{j} + (3 - (-3))\hat{k} = 3\hat{i} - 2\hat{j} + 6\hat{k}$.
The plane passes through the point $Q(1, -3, 3)$.
The equation of the plane passing through $(x_0, y_0, z_0)$ with normal $\vec{n} = a\hat{i} + b\hat{j} + c\hat{k}$ is $a(x - x_0) + b(y - y_0) + c(z - z_0) = 0$.
Substituting the values: $3(x - 1) - 2(y + 3) + 6(z - 3) = 0$.
$3x - 3 - 2y - 6 + 6z - 18 = 0$.
$3x - 2y + 6z - 27 = 0$.