The equation of the plane passing through the points (2, –4, 0), (3,

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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 :

Options:

A . 125x – 90y – 79z = 340

B . 32x – 21y – 36z = 85

C . 73x + 61y – 22z = 85

D . 29x – 27y – 22z = 85

Answer: Option D
:
D
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)

∴

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} . . . (3)

∴

Normal to the plane (1) is perpendicular to the line (3).

∴

6a + 4b + 3c = 0 . . .(4)
From (2) and (4), we get

\frac{a}{- 29} = \frac{b}{27} = \frac{c}{22}

∴

Equation of the required plane is
– 29(x – 2) + 27(y + 1) + 22(z) = 0

\Rightarrow - 29 x + 27 y + 22 z + 85 = 0

\Rightarrow 29 x - 27 y - 22 z = 85

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