Question #40663

Let $\vec{p} = \left( \sum_{k=0}^2 \binom{2}{k}^2 \right) \hat{i} + \left( \sum_{k=0}^2 (-1)^k \binom{2}{k}^2 \right) \hat{j} + \binom{4}{2} \hat{k}$ and $\vec{q} = \left( \sum_{k=0}^2 \binom{2}{k} \binom{3}{k} \right) \hat{i} + \left( \sum_{k=0}^2 (-1)^k \binom{2}{k} \binom{3}{k} \right) \hat{j} + \binom{5}{2} \hat{k}$. Let $\vec{w} = (a - 2) \hat{i} + 4 \hat{j} + 3 \hat{k}$. It is given that the vectors $\vec{p}, \vec{q},$ and $\vec{w}$ are coplanar. A point $C$ with position vector $\vec{c}$ lies on the line passing through the points with position vectors $\vec{p}$ and $\vec{q}$, such that $\vec{c} = (1-\lambda)\vec{p} + \lambda\vec{q}$. If the scalar projection of $\vec{c}$ on the vector $\vec{v} = -\hat{i} + (a-1)\hat{j} - \hat{k}$ is $-12\sqrt{2}$, then the value of $2\lambda$ is: