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The two planes $2 x - y - 2 z - 6 = 0,$ $3 x + 2 y - λ z + 1 = 0$ are perpendicular to each other then the absolute value of $λ$ is

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answer is B.

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Detailed Solution

$λ + 2 = 0$

The condition that the two planes perpendicular is $a_{1} a_{2} + b_{1} b_{2} + c_{1} c_{2} = 0$

$6 - 2 + 2 λ = 0 2 λ + 4 = 0 λ = - 2$

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