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Question

# The equation of the plane through the point (0, –4, –6) and (–2, 9, 3) and perpendicular to the plane x – 4y – 2z = 8 is

A
3x + 3y – 2z = 0
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B
x – 2y + z = 2
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C
2x + y – z = 2
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D
5x – 3y + 2z = 0
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Solution

## The correct option is C 2x + y – z = 2Equation of plane through (0, –4, –6) is a(x – 0) + b(y + 4) + c(z + 6) = 0 . . . .(i) point (–2, 9, 3) lies on Eq. (i), then – 2a + 13b + 9c = 0 . . . .(ii) also required plane is perpendicular to x – 4y – 2z = 8 a – 4b – 2c = 0 . . . .(iii) From Eqs. (i), (ii), (iii) we get ∣∣ ∣∣xy−4z+6−21391−4−2∣∣ ∣∣=0 ie, 2x + y – z = 2

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