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If the lines $\frac{x-1}{-3}=\frac{y-2}{2 k}=\frac{z-3}{2}$ and $\frac{x-1}{3 k}=\frac{y-1}{1}=\frac{z-6}{-5}$ are perpendicular, find the value of $k$.
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Verified Answer
The two lines are
$$
\begin{aligned}
&\frac{x-1}{3}=\frac{y-2}{2 k}=\frac{z-3}{2} \\
&\frac{x-1}{3 k}=\frac{y-1}{1}=\frac{z-6}{-5}
\end{aligned}
$$
The direction ratios of the lines (i) and (ii) are $-3,2 \mathrm{k}, 2$ and $3 \mathrm{k}, 1,-5$ lines (i) and (ii) are perpendicular to each other if $\mathrm{a}_1 \mathrm{a}_2+\mathrm{b}_1 \mathrm{~b}_2+\mathrm{c}_1 \mathrm{c}_2=0 \Rightarrow-9 \mathrm{k}+2 \mathrm{k}-10=0$
$$
\therefore \quad \mathrm{k}=-\frac{10}{7}
$$
$$
\begin{aligned}
&\frac{x-1}{3}=\frac{y-2}{2 k}=\frac{z-3}{2} \\
&\frac{x-1}{3 k}=\frac{y-1}{1}=\frac{z-6}{-5}
\end{aligned}
$$
The direction ratios of the lines (i) and (ii) are $-3,2 \mathrm{k}, 2$ and $3 \mathrm{k}, 1,-5$ lines (i) and (ii) are perpendicular to each other if $\mathrm{a}_1 \mathrm{a}_2+\mathrm{b}_1 \mathrm{~b}_2+\mathrm{c}_1 \mathrm{c}_2=0 \Rightarrow-9 \mathrm{k}+2 \mathrm{k}-10=0$
$$
\therefore \quad \mathrm{k}=-\frac{10}{7}
$$
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