1. **State the problem:** Find the distance between point $C(1,2,0)$ and the line $p$ given by the parametric equation $$\mathbf{r}(t) = (3,0,1) + t(1,-1,3), \quad t \in \mathbb{R}.$$\n\n2. **Formula used:** The distance $d$ from a point $\mathbf{P}$ to a line defined by a point $\mathbf{A}$ on the line and direction vector $\mathbf{v}$ is given by:\n$$d = \frac{\| (\mathbf{P} - \mathbf{A}) \times \mathbf{v} \|}{\| \mathbf{v} \|}.$$\nHere, $\times$ denotes the cross product and $\| \cdot \|$ denotes the vector magnitude.\n\n3. **Identify vectors:**\n- Point $\mathbf{P} = (1,2,0)$\n- Point on line $\mathbf{A} = (3,0,1)$\n- Direction vector $\mathbf{v} = (1,-1,3)$\n\n4. **Calculate vector $\mathbf{P} - \mathbf{A}$:**\n$$\mathbf{P} - \mathbf{A} = (1-3, 2-0, 0-1) = (-2, 2, -1).$$\n\n5. **Calculate cross product $(\mathbf{P} - \mathbf{A}) \times \mathbf{v}$:**\n$$\begin{aligned} (\mathbf{P} - \mathbf{A}) \times \mathbf{v} &= \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ -2 & 2 & -1 \\ 1 & -1 & 3 \end{vmatrix} \\ &= \mathbf{i}(2 \cdot 3 - (-1) \cdot (-1)) - \mathbf{j}(-2 \cdot 3 - (-1) \cdot 1) + \mathbf{k}(-2 \cdot (-1) - 2 \cdot 1) \\ &= \mathbf{i}(6 - 1) - \mathbf{j}(-6 + 1) + \mathbf{k}(2 - 2) \\ &= 5\mathbf{i} + 5\mathbf{j} + 0\mathbf{k} = (5,5,0). \end{aligned}$$\n\n6. **Calculate magnitudes:**\n$$\| (\mathbf{P} - \mathbf{A}) \times \mathbf{v} \| = \sqrt{5^2 + 5^2 + 0^2} = \sqrt{25 + 25} = \sqrt{50} = 5\sqrt{2}.$$\n$$\| \mathbf{v} \| = \sqrt{1^2 + (-1)^2 + 3^2} = \sqrt{1 + 1 + 9} = \sqrt{11}.$$\n\n7. **Calculate distance:**\n$$d = \frac{5\sqrt{2}}{\sqrt{11}} = \frac{5\sqrt{22}}{11}.$$\n\n**Final answer:** The distance from point $C$ to line $p$ is $$\boxed{\frac{5\sqrt{22}}{11}}.$$