Question:

Let $\vec{v} = 15\hat{i} + 10\hat{j} + 6\hat{k}$. We take the dot product of both sides with $\vec{p} \times \vec{q}$: $$\vec{v} \cdot (\vec{p} \times \vec{q}) = \alpha(2\vec{p} + \vec{q}) \cdot (\vec{p} \times \vec{q}) + \beta(\vec{p} - 2\vec{q}) \cdot (\vec{p} \times \vec{q}) + \gamma |\vec{p} \times \vec{q}|^2$$ Since $\vec{p} \times \vec{q}$ is orthogonal to both $\vec{p}$ and $\vec{q}$: $$\vec{v} \cdot (\vec{p} \times \vec{q}) = \gamma |\vec{p} \times \vec{q}|^2 \implies \gamma = \dfrac{\vec{v} \cdot (\vec{p} \times \vec{q})}{|\vec{p} \times \vec{q}|^2}$$ 1) Compute $\vec{p} \times \vec{q}$: $$\vec{p} \times \vec{q} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & 3 \\ 1 & -1 & 1 \end{vmatrix} = 4\hat{i} + \hat{j} - 3\hat{k}$$ 2) Compute magnitude squared: $$|\vec{p} \times \vec{q}|^2 = 4^2 + 1^2 + (-3)^2 = 16 + 1 + 9 = 26$$ 3) Compute $\vec{v} \cdot (\vec{p} \times \vec{q})$: $$\vec{v} \cdot (\vec{p} \times \vec{q}) = (15)(4) + (10)(1) + (6)(-3) = 60 + 10 - 18 = 52$$ 4) Solve for $\gamma$: $$\gamma = \dfrac{52}{26} = 2$$ Thus, the answer is 2.