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(C) Step $1$: Find two vectors in the plane. Let $\vec{AB} = (2-1)\hat{i} + (-1+2)\hat{j} + (-3-1)\hat{k} = \hat{i} + \hat{j} - 4\hat{k}$ and $\vec{AC

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$\frac{5}{\sqrt{17}}$

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(C) Step $1$: Find two vectors in the plane. Let $\vec{AB} = (2-1)\hat{i} + (-1+2)\hat{j} + (-3-1)\hat{k} = \hat{i} + \hat{j} - 4\hat{k}$ and $\vec{AC} = (0-1)\hat{i} + (1+2)\hat{j} + (5-1)\hat{k} = -\hat{i} + 3\hat{j} + 4\hat{k}$.Step $2$: Find the normal vector $\vec{n} = \vec{AB} 	imes \vec{AC} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & 1 & -4 \ -1 & 3 & 4 \end{vmatrix} = \hat{i}(4+12) - \hat{j}(4-4) + \hat{k}(3+1) = 16\hat{i} + 0\hat{j} + 4\hat{k}$.Step $3$: Simplify the normal vector to $\vec{n}' = 4\hat{i} + \hat{k}$.Step $4$: The equation of the plane is $4(x-1) + 0(y+2) + 1(z-1) = 0$, which simplifies to $4x + z - 5 = 0$.Step $5$: The perpendicular distance from $(0,0,0)$ to $Ax+By+Cz+D=0$ is $d = \frac{|D|}{\sqrt{A^2+B^2+C^2}} = \frac{|-5|}{\sqrt{4^2+0^2+1^2}} = \frac{5}{\sqrt{17}}$.

The perpendicular distance from the origin to the plane containing the points $A(1, -2, 1)$, $B(2, -1, -3)$ and $C(0, 1, 5)$ is (in units)

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