(a) Find the equation of the line passing through the point of intersection of the lines $\frac{x}{1} = \frac{y - 1}{2} = \frac{z - 2}{3}$ and $\frac{x - 1}{0} = \frac{y}{- 3} = \frac{z - 7}{2}$ and perpendicular to these given lines. OR (b) Two vertices of the parallelogram $ABCD$ are given as $A(- 1, 2, 1)$ and $B(1, - 2, 5)$. If the equation of the line passing through $C$ and $D$ is $\frac{x - 4}{1} = \frac{y + 7}{- 2} = \frac{z - 8}{2}$, then find the distance between sides $AB$ and $CD$. Hence, find the area of parallelogram $ABCD$.