Question

The equation of the plane passing through the points (2, –4, 0), (3, –4, 5) and parallel to the line 2x = 3y = 4z is :

• A. 125x – 90y – 79z = 340
• B. 32x – 21y – 36z = 85
• C. 73x + 61y – 22z = 85
• D. 29x – 27y – 22z = 85

Answer 1

The correct option is D 29x – 27y – 22z = 85

Equation of the plane through the point (2, –1, 0) is
a(x – 2) + b(y + 1) + c(z – 0) = 0 . . . (1)
It passes through the point (3, –4, 5)
$\therefore$ a – 3b + 5c = 0 . . . (2)
The plane (1) is parallel to the line
2x = 3y = 4z i.e.,$\frac{x}{6}=\frac{y}{4}=\frac{z}{3}...\left(3\right)$
$\therefore$ Normal to the plane (1) is perpendicular to the line (3).
$\therefore$ 6a + 4b + 3c = 0 . . .(4)
From (2) and (4), we get
$\frac{a}{-29}=\frac{b}{27}=\frac{c}{22}$
$\therefore$ Equation of the required plane is
– 29(x – 2) + 27(y + 1) + 22(z) = 0
$\Rightarrow -29x+27y+22z+85=0$
$\Rightarrow 29x-27y-22z=85$