Two lines \(\frac{x-3}{1}=\frac{y+1}{3}=\frac{z-6}{-1} \quad\) and \(\frac{x+5}{7}=\frac{y-2}{-6}=\frac{z-3}{4} \quad\) intersect at the point \(\mathrm{R}\). Then reflection of \(\mathrm{R}\) in the \(x y\)-plane has co-ordinates

Let \(\frac{x-3}{1}=\frac{y+1}{3}=\frac{z-6}{-1}=\lambda\)
\(\Rightarrow x=3+\lambda, y=3 \lambda-1, \mathrm{z}=-\lambda+6\)
Let \(\frac{x+5}{7}=\frac{y-2}{-6}=\frac{z-3}{4}=\mu\)
\(\Rightarrow x=7 \mu-5, y=-6 \mu+2, \mathrm{z}=4 \mu+3\)
Both the given lines intersect each other.
\(\begin{aligned}
& \text { So, } \lambda+3=7 \mu-5 \\
& \Rightarrow 7 \mu-\lambda=8...(i)
\end{aligned}\)
Also, \(3 \lambda-1=-6 \mu+2\)
\(\Rightarrow 6 \mu+3 \lambda=3\)...(ii)
From (i) and (ii), we get
\(\begin{aligned}
& \mu=1, \lambda=-1 \\
& \text { i.e., } x=2, y=-4, \mathrm{z}=7
\end{aligned}\)
\(\therefore \quad\) Co-ordinates of the intersection of the given lines are \(\mathrm{R}(2,-4,7)\)
Hence, reflection of \(\mathrm{R}\) in the \(x y\)-plane is \((2,-4,-7)\).