Let a vector $$\vec{a}$$ be coplanar with vectors $$\vec{b} = 2\hat{i} + \hat{j} + \hat{k}$$ and $$\vec{c} = \hat{i} - \hat{j} + \hat{k}$$. If $$\vec{a}$$ is perpendicular to $$\vec{d} = 3\hat{i} + 2\hat{j} + 6\hat{k}$$, and $$|\vec{a}| = \sqrt{10}$$. Then a possible value of $$[\vec{a} \ \vec{b} \ \vec{c}] + [\vec{a} \ \vec{b} \ \vec{d}] + [\vec{a} \ \vec{c} \ \vec{d}]$$ is equal to:

Since $$\vec{a}$$ is coplanar with $$\vec{b} = 2\hat{i} + \hat{j} + \hat{k}$$ and $$\vec{c} = \hat{i} - \hat{j} + \hat{k}$$, we can write $$\vec{a} = s\vec{b} + t\vec{c} = (2s + t)\hat{i} + (s - t)\hat{j} + (s + t)\hat{k}$$ for some scalars $$s, t$$.

The condition $$\vec{a} \perp \vec{d}$$ where $$\vec{d} = 3\hat{i} + 2\hat{j} + 6\hat{k}$$ gives $$\vec{a} \cdot \vec{d} = 3(2s + t) + 2(s - t) + 6(s + t) = 14s + 7t = 0$$, so $$t = -2s$$.

Substituting back: $$\vec{a} = (2s - 2s)\hat{i} + (s + 2s)\hat{j} + (s - 2s)\hat{k} = 3s\,\hat{j} - s\,\hat{k}$$. The magnitude condition $$|\vec{a}| = \sqrt{10}$$ gives $$\sqrt{9s^2 + s^2} = |s|\sqrt{10} = \sqrt{10}$$, so $$|s| = 1$$.

Taking $$s = 1$$: $$\vec{a} = (0, 3, -1)$$. Since $$\vec{a}$$ is coplanar with $$\vec{b}$$ and $$\vec{c}$$, the scalar triple product $$[\vec{a}\;\vec{b}\;\vec{c}] = 0$$.

For $$[\vec{a}\;\vec{b}\;\vec{d}] = \vec{a} \cdot (\vec{b} \times \vec{d})$$, compute $$\vec{b} \times \vec{d} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & 1 \\ 3 & 2 & 6 \end{vmatrix} = (6 - 2)\hat{i} - (12 - 3)\hat{j} + (4 - 3)\hat{k} = 4\hat{i} - 9\hat{j} + \hat{k}$$. Then $$[\vec{a}\;\vec{b}\;\vec{d}] = (0)(4) + (3)(-9) + (-1)(1) = -28$$.

For $$[\vec{a}\;\vec{c}\;\vec{d}] = \vec{a} \cdot (\vec{c} \times \vec{d})$$, compute $$\vec{c} \times \vec{d} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -1 & 1 \\ 3 & 2 & 6 \end{vmatrix} = (-6 - 2)\hat{i} - (6 - 3)\hat{j} + (2 + 3)\hat{k} = -8\hat{i} - 3\hat{j} + 5\hat{k}$$. Then $$[\vec{a}\;\vec{c}\;\vec{d}] = (0)(-8) + (3)(-3) + (-1)(5) = -14$$.

The total is $$0 + (-28) + (-14) = -42$$.