If the lines vec{r} = hat{i} + hat{j} - hat{k | vedclass

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(C) Step $1$: Find $q$ using the collinearity condition. The vector $q\hat{i} - 2\hat{j} + \hat{k}$ is collinear to $4\hat{i} - 4\hat{j} + 2\hat{k}$,

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$p = 4, q = 2$

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(C) Step $1$: Find $q$ using the collinearity condition. The vector $q\hat{i} - 2\hat{j} + \hat{k}$ is collinear to $4\hat{i} - 4\hat{j} + 2\hat{k}$, so $\frac{q}{4} = \frac{-2}{-4} = \frac{1}{2}$. Thus, $q = 4 	imes \frac{1}{2} = 2$.Step $2$: The lines are $\vec{r} = (1, 1, -1) + \lambda(2, -2, 1)$ and $\vec{r} = (p, -3, 2) + \mu(1, -2, 2)$.Step $3$: For intersection, $(1+2\lambda, 1-2\lambda, -1+\lambda) = (p+\mu, -3-2\mu, 2+2\mu)$.Step $4$: Equating $y$ and $z$ coordinates: $1-2\lambda = -3-2\mu \implies 2\lambda - 2\mu = 4 \implies \lambda - \mu = 2$. Also, $-1+\lambda = 2+2\mu \implies \lambda - 2\mu = 3$.Step $5$: Solving these, $\mu = -1$ and $\lambda = 1$.Step $6$: Equating $x$ coordinates: $1+2(1) = p + (-1) \implies 3 = p - 1 \implies p = 4$. Thus, $p = 4, q = 2$.

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