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The equation of the plane through the line of intersection of the planes 2x+3y+4z–7=0, x + y + z–1 = 0 and perpendicular to the plane x–5y + 3z–2 = 0 is

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7x - y - 6z - 17 = 0

x-y - 6z - 27 = 0

x + 2y + 3z - 6 = 0

x + 2y + 3z – 3 = 0

answer is C.

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Detailed Solution

$2 x + 3 y + 4 z - 7 + K (x + y + z - 1) = 0$

Which is $⊥ r$ to x-5y+3z-2=0

$(2 + k) - 5 (3 + K) + 3 (4 + k) = 0 \Rightarrow K = - 1$

Required plane $x + 2 y + 3 z - 6 = 0$

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