If the solve the problem
Question:

Two lines $\quad \frac{x-3}{1}=\frac{y+1}{3}=\frac{z-6}{-1} \quad$ and

$\frac{x+5}{7}=\frac{y-2}{-6}=\frac{z-3}{4}$ intersect at the point $R$. The

reflection of $R$ in the $x y$-plane has coordinates :-

1. $(2,4,7)$

2. $(-2,4,7)$

3. $(2,-4,-7)$

4. $(2,-4,7)$

Correct Option: , 3

Solution:

Point on $\mathrm{L}_{1}(\lambda+3,3 \lambda-1,-\lambda+6)$

Point on $\mathrm{L}_{2}(7 \mu-5,-6 \mu+2,4 \mu+3)$

$\Rightarrow \lambda+3=7 \mu-5$    ……(i)

$3 \lambda-1=-6 \mu+2$            ……(ii) $\Rightarrow \lambda=-1, \mu=1$

point $R(2,-4,7)$

Reflection is $(2,-4,-7)$