If the line drawn from the point (–2, – 1, – 3)

Given, points (–2, – 1, – 3) and (1, – 3, 3)

Direction ratios of the normal to the plane are (1 + 2, -3 + 1, 3 + 3) = (3, -2, 6)

Now, the equation of plane passing through one point (x1, y1, z1) is

a(x – x1) + b(y – y1) + c(z – z1) = 0

3(x – 1) – 2(y + 3) + 6(z – 3) = 0

3x – 3 – 2y – 6 + 6z – 18 = 0

3x – 2y + 6z – 27 = 0 ⇒ 3x – 2y + 6z = 27

Thus, the required equation of plane is 3x – 2y + 6z = 27.