$\overrightarrow{AB} = 3\hat{i} - \hat{j} + \hat{k}$ and $\overrightarrow{CD} = -3\hat{i} + 2\hat{j} + 4\hat{k}$ are two vectors. The position vectors of the points $A$ and $C$ are $6\hat{i} + 7\hat{j} + 4\hat{k}$ and $-9\hat{j} + 2\hat{k}$ respectively. Find the position vectors of a point $P$ on the line $AB$ and a point $Q$ on the line $CD$ such that $\overrightarrow{PQ}$ is perpendicular to both $\overrightarrow{AB}$ and $\overrightarrow{CD}$.

(N/A) Let the line $AB$ be represented by $\vec{r} = (6\hat{i} + 7\hat{j} + 4\hat{k}) + \lambda(3\hat{i} - \hat{j} + \hat{k})$. Any point $P$ on $AB$ is $(6 + 3\lambda, 7 - \lambda, 4 + \lambda)$.
Let the line $CD$ be represented by $\vec{r} = (0\hat{i} - 9\hat{j} + 2\hat{k}) + \mu(-3\hat{i} + 2\hat{j} + 4\hat{k})$. Any point $Q$ on $CD$ is $(-3\mu, -9 + 2\mu, 2 + 4\mu)$.
Then $\overrightarrow{PQ} = (-3\mu - 6 - 3\lambda)\hat{i} + (2\mu + \lambda - 16)\hat{j} + (4\mu - \lambda - 2)\hat{k}$.
Since $\overrightarrow{PQ} \perp \overrightarrow{AB}$,the dot product is $0$: $3(-3\mu - 6 - 3\lambda) - 1(2\mu + \lambda - 16) + 1(4\mu - \lambda - 2) = 0 \Rightarrow -7\mu - 11\lambda - 4 = 0 \dots (i)$.
Since $\overrightarrow{PQ} \perp \overrightarrow{CD}$,the dot product is $0$: $-3(-3\mu - 6 - 3\lambda) + 2(2\mu + \lambda - 16) + 4(4\mu - \lambda - 2) = 0 \Rightarrow 29\mu + 7\lambda - 22 = 0 \dots (ii)$.
Solving $(i)$ and $(ii)$,we multiply $(i)$ by $7$ and $(ii)$ by $11$: $-49\mu - 77\lambda - 28 = 0$ and $319\mu + 77\lambda - 242 = 0$.
Adding these,$270\mu - 270 = 0 \Rightarrow \mu = 1$. Substituting $\mu = 1$ in $(i)$,$-7 - 11\lambda - 4 = 0 \Rightarrow \lambda = -1$.
Thus,$P = (6 + 3(-1), 7 - (-1), 4 + (-1)) = (3, 8, 3)$ and $Q = (-3(1), -9 + 2(1), 2 + 4(1)) = (-3, -7, 6)$.
The position vectors are $\vec{P} = 3\hat{i} + 8\hat{j} + 3\hat{k}$ and $\vec{Q} = -3\hat{i} - 7\hat{j} + 6\hat{k}$.