Square of Area of Triangle Formed by Vector Combinations

To find the square of the area of the triangle formed by the vectors $2\vec{a} + 3\vec{b}$ and $\vec{a} - \vec{b}$, we first determine the cross product of the two adjacent sides.

The cross product of the given vectors is: $(2\vec{a} + 3\vec{b}) \times (\vec{a} - \vec{b}) = 2(\vec{a} \times \vec{a}) - 2(\vec{a} \times \vec{b}) + 3(\vec{b} \times \vec{a}) - 3(\vec{b} \times \vec{b})$

Since $\vec{a} \times \vec{a} = \vec{0}$, $\vec{b} \times \vec{b} = \vec{0}$, and $\vec{b} \times \vec{a} = -(\vec{a} \times \vec{b})$, this simplifies to: $(2\vec{a} + 3\vec{b}) \times (\vec{a} - \vec{b}) = -5(\vec{a} \times \vec{b})$

Next, we calculate the cross product $\vec{a} \times \vec{b}$ using the given vector components $\vec{a} = 2\hat{i} + 3\hat{j} + 3\hat{k}$ and $\vec{b} = 6\hat{i} + 3\hat{j} + 3\hat{k}$: $\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & 3 \\ 6 & 3 & 3 \end{vmatrix} = 0\hat{i} + 12\hat{j} - 12\hat{k}$

The magnitude squared of $\vec{a} \times \vec{b}$ is: $|\vec{a} \times \vec{b}|^2 = 0^2 + 12^2 + (-12)^2 = 288$

The area $A$ of the triangle is given by half the magnitude of the cross product of its adjacent sides: $A = \frac{1}{2} |(2\vec{a} + 3\vec{b}) \times (\vec{a} - \vec{b})| = \frac{5}{2} |\vec{a} \times \vec{b}|$

Squaring the area gives: $A^2 = \frac{25}{4} |\vec{a} \times \vec{b}|^2 = \frac{25}{4} \times 288 = 1800$

Thus, the square of the area of the triangle is $1800$, which corresponds to Option C.