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If two lines $L_1: \frac{x-1}{2}=\frac{y+1}{3}=\frac{z-1}{4}$ and $L_2: \frac{x-3}{1}=\frac{y-k}{2}=z$ intersect at a point, then $2 k$ is equal to
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The correct answer is:
9
Let $\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-1}{4}=\lambda$
Now, any point $P$ that lies on the lines $L_1$ has the form $(1+2 \lambda,-1+3 \lambda, 1+4 \lambda)$.
Now, on putting $x=1+2 \lambda, y=-1+3 \lambda$ and $z=1+4 \lambda$ into the equation of lines $L_2$, we get
$\begin{aligned}
& \frac{1+2 \lambda-3}{1}=\frac{-1+3 \lambda-k}{2}=1+4 \lambda \\
& \Rightarrow \quad \frac{1+2 \lambda-3}{1}=1+4 \lambda \\
& -2 \lambda=3 \Rightarrow \lambda=\frac{-3}{2} \\
& \frac{-1+3 \lambda-k}{2}=1+4 \lambda \\
& -1+3 \lambda-k=2+8 \lambda \\
& -5 \lambda=3+k \\
& -5\left(-\frac{3}{2}\right)=3+k \quad[\because \lambda=-3 / 2] \\
& k=\frac{15}{2}-3 \\
& k=\frac{9}{2} \\
& 2 k=9 \\
&
\end{aligned}$
Now, any point $P$ that lies on the lines $L_1$ has the form $(1+2 \lambda,-1+3 \lambda, 1+4 \lambda)$.
Now, on putting $x=1+2 \lambda, y=-1+3 \lambda$ and $z=1+4 \lambda$ into the equation of lines $L_2$, we get
$\begin{aligned}
& \frac{1+2 \lambda-3}{1}=\frac{-1+3 \lambda-k}{2}=1+4 \lambda \\
& \Rightarrow \quad \frac{1+2 \lambda-3}{1}=1+4 \lambda \\
& -2 \lambda=3 \Rightarrow \lambda=\frac{-3}{2} \\
& \frac{-1+3 \lambda-k}{2}=1+4 \lambda \\
& -1+3 \lambda-k=2+8 \lambda \\
& -5 \lambda=3+k \\
& -5\left(-\frac{3}{2}\right)=3+k \quad[\because \lambda=-3 / 2] \\
& k=\frac{15}{2}-3 \\
& k=\frac{9}{2} \\
& 2 k=9 \\
&
\end{aligned}$
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