Let $$\vec{a} = 2\hat{i} + 3\hat{j} + 3\hat{k}$$ and $$\vec{b} = 6\hat{i} + 3\hat{j} + 3\hat{k}$$. Then the square of the area of the triangle with adjacent sides determined by the vectors $$(2\vec{a} + 3\vec{b})$$ and $$(\vec{a} - \vec{b})$$ is :

$$\text{Area} = \frac{1}{2} |\vec{u} \times \vec{v}|$$

$$\vec{u} = 2\vec{a} + 3\vec{b}$$

$$\vec{v} = \vec{a} - \vec{b}$$

$$\vec{u} \times \vec{v} = (2\vec{a} + 3\vec{b}) \times (\vec{a} - \vec{b})$$

$$\vec{u} \times \vec{v} = 2(\vec{a} \times \vec{a}) - 2(\vec{a} \times \vec{b}) + 3(\vec{b} \times \vec{a}) - 3(\vec{b} \times \vec{b})$$

$$\vec{u} \times \vec{v} = \vec{0} - 2(\vec{a} \times \vec{b}) - 3(\vec{a} \times \vec{b}) - \vec{0}$$

$$\vec{u} \times \vec{v} = -5(\vec{a} \times \vec{b})$$

$$\vec{a} \times \vec{b} = \text{det} \begin{bmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & 3 \\ 6 & 3 & 3 \end{bmatrix}$$

$$\vec{a} \times \vec{b} = \hat{i}(3 \cdot 3 - 3 \cdot 3) - \hat{j}(2 \cdot 3 - 3 \cdot 6) + \hat{k}(2 \cdot 3 - 3 \cdot 6)$$

$$\vec{a} \times \vec{b} = \hat{i}(0) - \hat{j}(6 - 18) + \hat{k}(6 - 18)$$

$$\vec{a} \times \vec{b} = 12\hat{j} - 12\hat{k}$$

$$\vec{u} \times \vec{v} = -5(12\hat{j} - 12\hat{k}) = -60\hat{j} + 60\hat{k}$$

$$|\vec{u} \times \vec{v}|^2 = (-60)^2 + (60)^2 = 3600 + 3600 = 7200$$

$$\text{Area}^2 = \frac{1}{4} |\vec{u} \times \vec{v}|^2$$

$$\text{Area}^2 = \frac{1}{4} \cdot 7200 = 1800$$