Let A(2, 3, 0), B(0, 3, 2) and C(4, 0, 3) be | vedclass

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(B) The area of a triangle with vertices $A$, $B$, and $C$ is given by $\frac{1}{2} |\vec{AB} 	imes \vec{AC}|$.$\vec{AB} = (0-2)\hat{i} + (3-3)\hat{j

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$\frac{\sqrt{172}}{2} 	ext{ sq. units}$

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(B) The area of a triangle with vertices $A$, $B$, and $C$ is given by $\frac{1}{2} |\vec{AB} 	imes \vec{AC}|$.$\vec{AB} = (0-2)\hat{i} + (3-3)\hat{j} + (2-0)\hat{k} = -2\hat{i} + 0\hat{j} + 2\hat{k}$.$\vec{AC} = (4-2)\hat{i} + (0-3)\hat{j} + (3-0)\hat{k} = 2\hat{i} - 3\hat{j} + 3\hat{k}$.$\vec{AB} 	imes \vec{AC} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ -2 & 0 & 2 \ 2 & -3 & 3 \end{vmatrix} = \hat{i}(0 - (-6)) - \hat{j}(-6 - 4) + \hat{k}(6 - 0) = 6\hat{i} + 10\hat{j} + 6\hat{k}$.$|\vec{AB} 	imes \vec{AC}| = \sqrt{6^2 + 10^2 + 6^2} = \sqrt{36 + 100 + 36} = \sqrt{172}$.Area $= \frac{1}{2} \sqrt{172} 	ext{ sq. units}$.

Let $A(2, 3, 0)$, $B(0, 3, 2)$ and $C(4, 0, 3)$ be the vertices of a triangle. Find the area of the triangle.

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