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

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answer is -8.

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Detailed Solution

If a plane contains the line, then its normal and line are perpendicular ∴ 4(2)+k(4)+8(3)=0(i.e., al+bm+cn=0) 8+4k+24=0⇒k=−8

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