Using vectors,find the area of the triangle $ABC$ with vertices $A(1, 2, 3)$,$B(2, -1, 4)$ and $C(4, 5, -1)$.

(N/A) The position vectors of the vertices are $\vec{A} = \hat{i} + 2\hat{j} + 3\hat{k}$,$\vec{B} = 2\hat{i} - \hat{j} + 4\hat{k}$,and $\vec{C} = 4\hat{i} + 5\hat{j} - \hat{k}$.
First,we find the vectors $\vec{AB}$ and $\vec{AC}$:
$\vec{AB} = \vec{B} - \vec{A} = (2-1)\hat{i} + (-1-2)\hat{j} + (4-3)\hat{k} = \hat{i} - 3\hat{j} + \hat{k}$
$\vec{AC} = \vec{C} - \vec{A} = (4-1)\hat{i} + (5-2)\hat{j} + (-1-3)\hat{k} = 3\hat{i} + 3\hat{j} - 4\hat{k}$
Now,calculate the cross product $\vec{AB} \times \vec{AC}$:
$\vec{AB} \times \vec{AC} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -3 & 1 \\ 3 & 3 & -4 \end{vmatrix}$
$= \hat{i}((-3)(-4) - (1)(3)) - \hat{j}((1)(-4) - (1)(3)) + \hat{k}((1)(3) - (-3)(3))$
$= \hat{i}(12 - 3) - \hat{j}(-4 - 3) + \hat{k}(3 + 9)$
$= 9\hat{i} + 7\hat{j} + 12\hat{k}$
Next,find the magnitude of the cross product:
$|\vec{AB} \times \vec{AC}| = \sqrt{9^2 + 7^2 + 12^2} = \sqrt{81 + 49 + 144} = \sqrt{274}$
The area of triangle $ABC$ is given by $\frac{1}{2} |\vec{AB} \times \vec{AC}|$:
Area $= \frac{1}{2} \sqrt{274} \text{ sq units}$.