(a) Find the distance between the line x/2=(2 y -6)/4=(1- z)/-1 and another line parallel to it passing through the point (4,0,-5).

Equation of the given line in standard form is $$\mathrm{L}_{1}: \frac{x}{2}=\frac{\mathrm{y}-3}{2}=\frac{\mathrm{z}-1}{1} $$ Equation of the line parallel to $\displaystyle \mathrm{L}_{1}$ & passing through ( $\displaystyle 4,0,-5$ ) is $$\mathrm{L}_{2}: \frac{x-4}{2}=\frac{\mathrm{y}}{2}=\frac{\mathrm{z}+5}{1} $$ Vector Equation of Lines are $\displaystyle \mathrm{L}_{1}: \vec{r}=(0 \hat{i}+3 \hat{j}+\hat{\mathrm{k}})+\lambda(2 \hat{i}+2 \hat{j}+\hat{\mathrm{k}})$ $$\mathrm{L}_{2}: \vec{r}=(4 \hat{i}+0 \hat{j}-5 \hat{\mathrm{k}})+\mu(2 \hat{i}+2 \hat{j}+\hat{\mathrm{k}}) $$ Now, $\displaystyle \overrightarrow{a_{2}}-\overrightarrow{a_{1}}=(4 \hat{i}+0 \hat{j}-5 \hat{\mathrm{k}})-(0 \hat{i}+3 \hat{j}+\hat{\mathrm{k}})=(4 \hat{i}-3 \hat{j}-6 \hat{\mathrm{k}})$ $$\begin{aligned} & \vec{b}=2 \hat{i}+2 \hat{j}+\hat{\mathrm{k}} \\ & \left(\overrightarrow{a_{2}}-\overrightarrow{a_{1}}\right) \times \vec{b}=\left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{\mathrm{k}} \\ 4 & -3 & -6 \\ 2 & 2 & 1 \end{array}\right|=9 \hat{i}-16 \hat{j}+14 \hat{\mathrm{k}} \end{aligned} $$ $$|\vec{b}|=\sqrt{4+4+1}=3 $$ Thus, distance between the lines is $$\text { S.D. }=\frac{\left|\left(\overrightarrow{a_{2}}-\overrightarrow{a_{1}}\right) \times \vec{b}\right|}{|\vec{b}|}=\frac{\sqrt{81+256+196}}{3}=\frac{\sqrt{533}}{3} \text { units } $$