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The point of intersection of the lines $\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}$ and $\frac{x-4}{5}=\frac{y-1}{2}=z$ is
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Verified Answer
The correct answer is:
$(-1,-1,-1)$
Given lines are
$$
\begin{aligned}
&\frac{x-1}{2}=\frac{y-2}{3}=\frac{z}{4} \\
&\frac{x-4}{5}=\frac{y-1}{2}=z
\end{aligned}
$$
$[$ say $] \ldots($ i)
and $\quad \frac{x-4}{5}=\frac{y-1}{2}=z$ Any point on the line (i) is $(2 r+1,3 r+2,4 r+3)$. If they intersect, then this point satisfies the second line, we get $$ \frac{2 r+1-4}{5}=\frac{3 r+2-1}{2}=4 r+3 $$
$\Rightarrow \quad \frac{2 r-3}{5}=\frac{3 r+1}{2} \Rightarrow r=-1$
Hence, the required point is $(-1,-1,-1)$.
$$
\begin{aligned}
&\frac{x-1}{2}=\frac{y-2}{3}=\frac{z}{4} \\
&\frac{x-4}{5}=\frac{y-1}{2}=z
\end{aligned}
$$
$[$ say $] \ldots($ i)
and $\quad \frac{x-4}{5}=\frac{y-1}{2}=z$ Any point on the line (i) is $(2 r+1,3 r+2,4 r+3)$. If they intersect, then this point satisfies the second line, we get $$ \frac{2 r+1-4}{5}=\frac{3 r+2-1}{2}=4 r+3 $$
$\Rightarrow \quad \frac{2 r-3}{5}=\frac{3 r+1}{2} \Rightarrow r=-1$
Hence, the required point is $(-1,-1,-1)$.
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