If the lines $\vec{r} = \hat{i} + \hat{j} - \hat{k} + \lambda(q\hat{i} - 2\hat{j} + \hat{k})$ and $\vec{r} = p\hat{i} - 3\hat{j} + 2\hat{k} + \mu(\hat{i} - 2\hat{j} + 2\hat{k})$ intersect each other and $q\hat{i} - 2\hat{j} + \hat{k}$ is collinear to $4\hat{i} - 4\hat{j} + 2\hat{k}$, then the values of $p$ and $q$ are

Given that the vector $q\hat{i} - 2\hat{j} + \hat{k}$ is collinear to $4\hat{i} - 4\hat{j} + 2\hat{k}$, their direction ratios must be proportional.

$\dfrac{q}{4} = \dfrac{-2}{-4} = \dfrac{1}{2}$

$\Rightarrow q = 2$

The equations of the lines become:

$\vec{r} = \hat{i} + \hat{j} - \hat{k} + \lambda(2\hat{i} - 2\hat{j} + \hat{k})$

$\vec{r} = p\hat{i} - 3\hat{j} + 2\hat{k} + \mu(\hat{i} - 2\hat{j} + 2\hat{k})$

Since the lines intersect, their general points must coincide for some values of $\lambda$ and $\mu$. Equating the respective components, we get:

$1 + 2\lambda = p + \mu$

$1 - 2\lambda = -3 - 2\mu$

$-1 + \lambda = 2 + 2\mu$

From the second equation:

$2\lambda - 2\mu = 4 \Rightarrow \lambda - \mu = 2$

From the third equation:

$\lambda - 2\mu = 3$

Subtracting the third equation from the modified second equation:

$(\lambda - \mu) - (\lambda - 2\mu) = 2 - 3$

$\mu = -1$

Substituting $\mu = -1$ into $\lambda - \mu = 2$:

$\lambda - (-1) = 2 \Rightarrow \lambda = 1$

Substituting $\lambda = 1$ and $\mu = -1$ into the first equation:

$1 + 2(1) = p - 1$

$3 = p - 1 \Rightarrow p = 4$

Thus, $p = 4$ and $q = 2$.

Answer: $p = 4,\ q = 2$