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The lines $\frac{x-1}{2}=\frac{y-4}{4}=\frac{z-2}{3}$ and $\frac{1-x}{1}=\frac{y-2}{5}=\frac{3-z}{a}$ are perpendicular to each other, then $a$ equals to
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6
Let $L_1: \frac{x-1}{2}=\frac{y-4}{4}=\frac{z-2}{3}$
and $L_2=\frac{1-x}{1}=\frac{y-2}{5}=\frac{3-z}{a}$
the line $L_2$ can be written as $\frac{x-1}{-1}=\frac{y-2}{5}=\frac{z-3}{-a}$
Now, the DR's of lines $L_1$ and $L_2$ are $(2,4,3)$ and $(-1,5,-a)$ respectively.
Since, $L_1$ and $L_2$ are perpendicular to each other.
$\begin{array}{lrl}
\therefore & 2(-1)+4)(5)+3(-a)=0 \\
\Rightarrow & -2+20-3 a=0 \\
\Rightarrow & -3 a=-18 \Rightarrow a=6
\end{array}$
and $L_2=\frac{1-x}{1}=\frac{y-2}{5}=\frac{3-z}{a}$
the line $L_2$ can be written as $\frac{x-1}{-1}=\frac{y-2}{5}=\frac{z-3}{-a}$
Now, the DR's of lines $L_1$ and $L_2$ are $(2,4,3)$ and $(-1,5,-a)$ respectively.
Since, $L_1$ and $L_2$ are perpendicular to each other.
$\begin{array}{lrl}
\therefore & 2(-1)+4)(5)+3(-a)=0 \\
\Rightarrow & -2+20-3 a=0 \\
\Rightarrow & -3 a=-18 \Rightarrow a=6
\end{array}$
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