The three vectors $\vec{A}=3 \hat{i}-2 \hat{j}+\hat{k}, \vec{B}=\hat{i}-3 \hat{j}+5 \hat{k}$ and $\vec{C}=2 \hat{i}-\hat{j}+4 \hat{k}$ will form

Given vectors are $\mathrm{A} = 3 \hat{i} - 2 \hat{j} + \hat{k}$, $\mathrm{B} = \hat{i} - 3 \hat{j} + 5 \hat{k}$, and $\mathrm{C} = 2 \hat{i} - \hat{j} + 4 \hat{k}$. Magnitudes are: $|\mathrm{A}| = \sqrt{14}$, $|\mathrm{B}| = \sqrt{35}$, and $|\mathrm{C}| = \sqrt{21}$. All pairs satisfy the triangle inequality, confirming a valid triangle.

The squared magnitudes are: $|\mathrm{A}|^2 = 14$, $|\mathrm{B}|^2 = 35$, $|\mathrm{C}|^2 = 21$. Since $14 + 21 = 35$, the Pythagorean theorem holds: $|\mathrm{A}|^2 + |\mathrm{C}|^2 = |\mathrm{B}|^2$, indicating a right-angled triangle.

The triangle is not equilateral or isosceles due to unequal side lengths. Conclusion: Right-angled triangle. Answer: $\boxed{D}$