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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 :

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

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)
a – 3b + 5c = 0 . . . (2)
The plane (1) is parallel to the line
2x = 3y = 4z i.e.,x6=y4=z3...(3)
Normal to the plane (1) is perpendicular to the line (3).
6a + 4b + 3c = 0 . . .(4)
From (2) and (4), we get
a29=b27=c22
Equation of the required plane is
– 29(x – 2) + 27(y + 1) + 22(z) = 0
29x+27y+22z+85=0
29x27y22z=85



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