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If the lines $\frac{x-1}{-3}=\frac{y-2}{2 k}=\frac{z-3}{2}, \frac{x-1}{3 k}=\frac{y-5}{1}=\frac{z-6}{-5}$ are at right angles, then $k=$
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Verified Answer
The correct answer is:
$\frac{-10}{7}$
Lines are perpendicular if $a_1 a_2+b_1 b_2+c_1 c_2=0$
Hence,
$-3(3 k)+2 k(1)+2(-5)=0 \Rightarrow k=-\frac{10}{7} \text {. }$
Hence,
$-3(3 k)+2 k(1)+2(-5)=0 \Rightarrow k=-\frac{10}{7} \text {. }$
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