If the plane $$4(x$$−$$1)+k(y$$−$$2)+8(z$$−$$5)=0$$ contains the line x−$$12=y$$−$$24=z$$−$$53$$, then k is

Planes in 3D

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Question

If the plane $4 (x - 1) + k (y - 2) + 8 (z - 5) = 0$ contains the line $\frac{x - 1}{2} = \frac{y - 2}{4} = \frac{z - 5}{3}$ , then $k$ is

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Solution

If a plane contains the line, then its normal and line are perpendicular
$∴ 4 (2) + k (4) + 8 (3) = 0$
$(i . e ., a l + b m + c n = 0)$
$8 + 4 k + 24 = 0$ $\Rightarrow k = - 8$

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