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(C) Step $1$: Find the normal vector $\vec{n}$ to the plane. The plane contains the line with direction $\vec{b} = 2\hat{i} - 2\hat{j} + \hat{k}$ and

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$5$

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(C) Step $1$: Find the normal vector $\vec{n}$ to the plane. The plane contains the line with direction $\vec{b} = 2\hat{i} - 2\hat{j} + \hat{k}$ and passes through $A(1, -3, 1)$ and $Q(1, 0, -1)$.Step $2$: The vector $\vec{AQ} = (1-1)\hat{i} + (0-(-3))\hat{j} + (-1-1)\hat{k} = 3\hat{j} - 2\hat{k}$.Step $3$: $\vec{n} = \vec{b} 	imes \vec{AQ} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 2 & -2 & 1 \ 0 & 3 & -2 \end{vmatrix} = \hat{i}(4-3) - \hat{j}(-4-0) + \hat{k}(6-0) = \hat{i} + 4\hat{j} + 6\hat{k}$.Step $4$: The equation of the plane is $1(x-1) + 4(y-0) + 6(z+1) = 0$, which simplifies to $x + 4y + 6z + 5 = 0$.Step $5$: The perpendicular distance from the origin $(0,0,0)$ is $d = \frac{|0 + 4(0) + 6(0) + 5|}{\sqrt{1^2 + 4^2 + 6^2}} = \frac{5}{\sqrt{1 + 16 + 36}} = \frac{5}{\sqrt{53}}$.Step $6$: Comparing with $\frac{p}{\sqrt{53}}$, we get $p = 5$.

If the perpendicular distance of the plane passing through the point $Q(1, 0, -1)$ and containing the line $\vec{r} = (\hat{i} - 3\hat{j} + \hat{k}) + \lambda(2\hat{i} - 2\hat{j} + \hat{k})$ from the origin is $\frac{p}{\sqrt{53}}$, then $p = $

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