If $A B ⊥ B C$ , then the value of $k$ is if $A (k, 1, - 1), B (2 k, 0, 2), C (2 + 2 k, k, 1)$

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Solution

The correct option is D $3$
Given, $A (k, 1, - 1), B (2 k, 0, 2), C (2 + 2 k, k, 1)$
$- - \to A B = (k, - 1, 3) - - \to B C = (2, k, - 1)$
$A B$ is perpendicular to $B C$
So, $- - \to A B . - - \to B C = 0$
$2 k - k - 3 = 0$
$∴ k = 3$

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