(a)

Find the shortest distance between the lines: \[\begin{aligned} & \frac{x+1}{2}=\frac{y-1}{1}=\frac{z-9}{-3} \text { and } \\ & \frac{x-3}{2}=\frac{y+15}{-7}=\frac{z-9}{5} \\ & \text { OR } \end{aligned} \]

(b)

Find the image $\displaystyle \mathrm{A}^{\prime}$ of the point $\displaystyle \mathrm{A}(2,1,2)$ in the line $\displaystyle l: \overrightarrow{\mathrm{r}}=4 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}+2 \hat{\mathrm{k}}+\lambda(\hat{\mathrm{i}}-\hat{\mathrm{j}}-\hat{\mathrm{k}})$. Also, find the equation of line joining $\displaystyle \mathrm{AA}^{\prime}$. Find the foot of perpendicular from point A on the line $\displaystyle l$.

The vector equations of the lines are \[\begin{aligned} & \vec{\mathrm{r}}=-\hat{\imath}+\hat{\jmath}+9 \hat{\mathrm{k}}+\lambda(2 \hat{\imath}+\hat{\jmath}-3 \hat{\mathrm{k}}) \\ & \vec{\mathrm{r}}=3 \hat{\imath}-15 \hat{\jmath}+9 \hat{\mathrm{k}}+\mu(2 \hat{\imath}-7 \hat{\jmath}+5 \hat{\mathrm{k}}) \\ & \overrightarrow{a_{1}}=-\hat{\imath}+\hat{\jmath}+9 \hat{\mathrm{k}}, \overrightarrow{a_{2}}=3 \hat{\imath}-15 \hat{\jmath}+9 \hat{\mathrm{k}} \\ & \overrightarrow{b_{1}}=2 \hat{\imath}+\hat{\jmath}-3 \hat{\mathrm{k}}, \overrightarrow{b_{2}}=2 \hat{\imath}-7 \hat{\jmath}+5 \hat{\mathrm{k}} \\ & \overrightarrow{a_{2}}-\overrightarrow{a_{1}}=4 \hat{\imath}-16 \hat{\jmath} \end{aligned} \] Let the image of A in the line be $\displaystyle \mathrm{A}^{\prime}(\alpha, \beta, \gamma)$ The point P , which is the point of intersection of the lines $\displaystyle l$ and $\displaystyle \mathrm{A} \mathrm{A}^{\prime}$, will have coordinates ( $\displaystyle \lambda+4,-\lambda+2,-\lambda+2$ ) for some $\displaystyle \lambda$. Drs of AP are $\displaystyle <\lambda+2,-\lambda+1,-\lambda>$ $\displaystyle \mathrm{A} P \perp l$ \[\begin{aligned} & (\lambda+2)-(-\lambda+1)-(-\lambda)=0 \\ & \Rightarrow \lambda=-\frac{1}{3} \end{aligned} \] Therefore, the coordinates of P are $\displaystyle \left(\frac{11}{3}, \frac{7}{3}, \frac{7}{3}\right)$ P is the mid-point of $\displaystyle \mathrm{A} \mathrm{A}^{\prime}$ \[\begin{aligned} & \Rightarrow \frac{2+\alpha}{2}=\frac{11}{3}, \frac{1+\beta}{2}=\frac{7}{3}, \frac{2+\gamma}{2}=\frac{7}{3} \\ & \Rightarrow \alpha=\frac{16}{3}, \beta=\frac{11}{3}, \gamma=\frac{8}{3} \end{aligned} \] The coordinates of the image are $\displaystyle \left(\frac{16}{3}, \frac{11}{3}, \frac{8}{3}\right)$ The equation of $\displaystyle \mathrm{A} \mathrm{A}^{\prime}$ is