(a) Find the point Q on the line (2 x +4)/6=(y +1)/2=(-2 z +6)/-4 at a distance of 3 √ 2 from the point P (1,2,3).

The general point on the line $\displaystyle (3 \lambda-2,2 \lambda-1,2 \lambda+3)$ is $\displaystyle \mathbf{Q}$, from some $\displaystyle \lambda \in \mathbf{R}$

\[\mathrm{P} \mathrm{Q}=3 \sqrt{2} \Rightarrow(\mathrm{P} \mathrm{Q})^{2}=18 \Rightarrow(3 \lambda-3)^{2}+(2 \lambda-3)^{2}+(2 \lambda)^{2}=18 \]

Thus, the point is $\displaystyle \mathrm{Q}(-2,-1,3)$ or $\displaystyle \mathrm{Q}\left(\frac{56}{17}, \frac{43}{17}, \frac{111}{17}\right)$

Let $\displaystyle \mathbf{A}^{\prime}(\mathbf{a}, \mathbf{b}, \mathbf{c})$ be the image of the point $\displaystyle \mathbf{A}(-\mathbf{1}, \mathbf{5}, \mathbf{2})$ in the given line, also assume ' $\displaystyle M$ ' as the point of intersection of $\displaystyle A A^{\prime}$ with the given line, then ' $\displaystyle M$ ' is the mid-point of the line segment $\displaystyle \mathbf{A} \mathbf{A}^{\prime}$

And, $\displaystyle \mathbf{A A}^{\prime}=\sqrt{\mathbf{4 9 + 6 4 + 9}}=\sqrt{\mathbf{1 2 2}}$