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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 = 2
Equation 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
∣ ∣xy4z+62139142∣ ∣=0
ie, 2x + y – z = 2

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