Question

The equation of the plane passing through the point (–2, –2, 2) and containing the line joining the points (1, 1, 1) and (1, –1, 2) is :

• A. x + 2y – 3z + 4 = 0
• B. 3z – 4y + 1 = 0
• C. 5x + 2y – 3z – 17 = 0
• D. x – 3y – 6z + 8 = 0

Answer 1

The correct option is D x – 3y – 6z + 8 = 0

Equation of a plane through (–2, –2, 2) is given by a(x + 2) + b(y + 2) + c(z – 2) = 0
It contains the line joining the points (1, 1, 1) and (1, –1, 2), so these points also lie in the plane.
$\therefore$ 3a + 3b – c = 0,
3a + b + 0.c = 0
$\therefore \frac{a}{1}=\frac{b}{-3}=\frac{c}{-6}=r$ (say)
or a = r, b = - 3r, c = - 6r
$\therefore$ Equation of the plane is x - 3y - 6z + 8 = 0