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 a−29=b27=c22 ∴ Equation of the required plane is – 29(x – 2) + 27(y + 1) + 22(z) = 0 ⇒−29x+27y+22z+85=0 ⇒29x−27y−22z=85