(a) Find the image A^′ of the point A (1,6,3) in the line x/1=(y -1)/2=(z -2)/3. Also, find the equation of the line joining A and A^′.

The equation of given line is $\displaystyle \frac{x}{1}=\frac{\mathrm{y}-1}{2}=\frac{\mathrm{z}-2}{3}=\lambda$ Any arbitrary point on the line is $\displaystyle M(\lambda, 2 \lambda+1,3 \lambda+2)$ dr's of $\displaystyle \boldsymbol{\mathrm{A}} \boldsymbol{M}$ are $\displaystyle <\boldsymbol{\lambda}-\mathbf{1}, 2 \boldsymbol{\lambda}-\mathbf{5}, 3 \boldsymbol{\lambda}-\mathbf{1}>$ Here1 $\displaystyle (\lambda-1)+2(2 \lambda-5)+3(3 \lambda-1)=0$ $\displaystyle \Rightarrow \lambda=1$ $\displaystyle \therefore M(1,3,5)$ is the foot perpendicular of the point A to the given line. Let image of point $\displaystyle \mathbf{A}$ in the line be $\displaystyle \boldsymbol{\mathrm{A}}^{\boldsymbol{\prime}} \boldsymbol{(} \boldsymbol{\alpha} \boldsymbol{,} \boldsymbol{\beta} \boldsymbol{,} \boldsymbol{\gamma} \boldsymbol{)}$ Since $\displaystyle M$ is the mid-point of $\displaystyle \mathrm{A} \mathrm{A}^{\prime}$, so $\displaystyle M\left(\frac{1+\alpha}{2}, \frac{6+\beta}{2}, \frac{3+\gamma}{2}\right)=M(1,3,5)$ $\displaystyle \Rightarrow \mathrm{A}^{\prime}(1,0,7)$ is the image of $\displaystyle \mathrm{A}$. Also, Equation of $\displaystyle \mathrm{A} \mathrm{A}^{\prime}$ is $\displaystyle \frac{x-1}{0}=\frac{\mathrm{y}-6}{-3}=\frac{\mathrm{z}-3}{2}$ The given line is $\displaystyle \frac{x+5}{1}=\frac{\mathrm{y}+3}{4}=\frac{\mathrm{z}-6}{-9}=\lambda$ and $\displaystyle \mathrm{Q}(2,4,-1)$ Any random point on the line will be given by $\displaystyle \mathrm{P}(\lambda-5,4 \lambda-3,-9 \lambda+6)$ Since $\displaystyle \boldsymbol{\mathrm{P} \mathrm{Q}}=7 \Rightarrow \sqrt{(\lambda-7)^{2}+(\mathbf{4} \lambda-7)^{2}+(-\mathbf{9} \lambda+7)^{2}}=7$ \[\Rightarrow 98\left(\lambda^{2}-2 \lambda+1\right)=0 \Rightarrow \lambda=1 \] Hence, the required point is $\displaystyle \boldsymbol{\mathrm{P}}(-\mathbf{4}, \mathbf{1},-\mathbf{3})$ The equation of line PQ is $\displaystyle \frac{x+4}{6}=\frac{\mathrm{y}-1}{3}=\frac{\mathrm{z}+3}{2}$ or $\displaystyle \frac{x-2}{6}=\frac{\mathrm{y}-4}{3}=\frac{\mathrm{z}+1}{2}$