Find the equation of a line $\displaystyle l_{2}$ which is the mirror image of the line $\displaystyle l_{1}$ with respect to line $\displaystyle l: \frac{\mathrm{x}}{1}=\frac{\mathrm{y}-1}{2}=\frac{\mathrm{z}-2}{3}$, given that line $\displaystyle l_{1}$ passes through the point $\displaystyle \mathrm{P}(1,6,3)$ and parallel to line $\displaystyle l$.

D ratios of the line $\displaystyle l$ i.e. $\displaystyle \dfrac{x}{1} = \dfrac{y-1}{2} = \dfrac{z-2}{3}$ are $\displaystyle 1,\ 2,\ 3$Let coordinates of foot of perpendicular M on line $\displaystyle l$ be $\displaystyle (\lambda,\ 2\lambda + 1,\ 3\lambda + 2)$D.ratios of PM are $\displaystyle \lambda - 1,\ 2\lambda - 5,\ 3\lambda - 1$$\displaystyle 1(\lambda - 1) + 2(2\lambda - 5) + 3(3\lambda - 1) = 0 \quad (\because \text{PM} \perp l)$$\displaystyle \Rightarrow \lambda = 1$Coordinates of M are $\displaystyle (1,\ 3,\ 5)$Since M is midpoint of PQ $\displaystyle \therefore$ Coordinates of Q are $\displaystyle (1,\ 0,\ 7)$Equation of line $\displaystyle l_2$ is $\displaystyle \dfrac{x-1}{1} = \dfrac{y}{2} = \dfrac{z-7}{3}$