1. **Problem Statement:** Find the volume of the tetrahedron determined by vectors $\mathbf{a} = 2\mathbf{i} - 3\mathbf{j} + \mathbf{k}$, $\mathbf{b} = \mathbf{i} + 2\mathbf{j} - \mathbf{k}$, and $\mathbf{c} = -3\mathbf{i} - \mathbf{j} + 5\mathbf{k}$.\n\n2. **Volume formula:** The volume $V$ of the tetrahedron formed by three vectors $\mathbf{a}, \mathbf{b}, \mathbf{c}$ is given by $$V = \frac{1}{6} |\mathbf{a} \cdot (\mathbf{b} \times \mathbf{c})|.$$\n\n3. **Calculate the cross product $\mathbf{b} \times \mathbf{c}$:**\n$$\mathbf{b} \times \mathbf{c} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & 2 & -1 \\ -3 & -1 & 5 \end{vmatrix} = \mathbf{i}(2 \cdot 5 - (-1) \cdot (-1)) - \mathbf{j}(1 \cdot 5 - (-1) \cdot (-3)) + \mathbf{k}(1 \cdot (-1) - 2 \cdot (-3)).$$\nSimplify inside the parentheses:\n$$= \mathbf{i}(10 - 1) - \mathbf{j}(5 - 3) + \mathbf{k}(-1 + 6) = 9\mathbf{i} - 2\mathbf{j} + 5\mathbf{k}.$$\n\n4. **Calculate the dot product $\mathbf{a} \cdot (\mathbf{b} \times \mathbf{c})$:**\n$$\mathbf{a} \cdot (\mathbf{b} \times \mathbf{c}) = (2)(9) + (-3)(-2) + (1)(5) = 18 + 6 + 5 = 29.$$\n\n5. **Calculate volume $V$:**\n$$V = \frac{1}{6} |29| = \frac{29}{6}.$$\n\n**Final answer:** The volume of the tetrahedron is $\boxed{\frac{29}{6}}$.