1. The problem asks to find the magnitude of the vector $\mathbf{v} = 3\mathbf{i} - 4\mathbf{j} + 12\mathbf{k}$.\n\n2. The magnitude (or length) of a vector $\mathbf{v} = a\mathbf{i} + b\mathbf{j} + c\mathbf{k}$ is given by the formula: $$|\mathbf{v}| = \sqrt{a^2 + b^2 + c^2}.$$\n\n3. Here, $a = 3$, $b = -4$, and $c = 12$. We square each component: $$3^2 = 9, \quad (-4)^2 = 16, \quad 12^2 = 144.$$\n\n4. Sum these squares: $$9 + 16 + 144 = 169.$$\n\n5. Take the square root of the sum: $$|\mathbf{v}| = \sqrt{169} = 13.$$\n\n6. Therefore, the magnitude of the vector $3\mathbf{i} - 4\mathbf{j} + 12\mathbf{k}$ is 13.