Question

The equation of the plane through the point (2, -1, -3) and parallel to the lines `(x - 1)/3 = (y + 2)/2 = z/(-4)` and `x/2 = (y - 1)/(-3) = (z - 2)/2` is ______

Options

• 8x + 14y + 13z + 37 = 0

• 8x - 14y + 13z + 37 = 0

• 8x + 14y - 13z + 37 = 0

• 8x + 14y + 13z - 37 = 0

Answer 1

The equation of the plane through the point (2, -1, -3) and parallel to the lines `(x - 1)/3 = (y + 2)/2 = z/(-4)` and `x/2 = (y - 1)/(-3) = (z - 2)/2` is 8x + 14y + 13z + 37 = 0.

Explanation:

The equation of plane passing through (2, -1, -3) is a(x - 2) + b(y + 1) + c(z + 3) = 0

Also, as the plane is parallel is the given two lines,

∴ 3a + 2b - 4c = 0 and 2a - 3b + 2c = 0

⇒ a = -8, b = -14, c = -13

∴ The equation of the required plane is

-8(x - 2) - 14(y + 1) - 13(z + 3) = 0

⇒ 8x + 14y + 13z + 37 = 0