For three vectors $$\vec{A} = (-x\hat{i} - 6\hat{j} - 2\hat{k})$$, $$\vec{B} = (-\hat{i} + 4\hat{j} + 3\hat{k})$$ and $$\vec{C} = (-8\hat{i} - \hat{j} + 3\hat{k})$$, if $$\vec{A} \cdot (\vec{B} \times \vec{C}) = 0$$, then value of $$x$$ is _________

$$\vec{A} \cdot (\vec{B} \times \vec{C}) = 0$$ (scalar triple product).

$$\vec{B} \times \vec{C} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -1 & 4 & 3 \\ -8 & -1 & 3 \end{vmatrix} = (12+3)\hat{i} - (-3+24)\hat{j} + (1+32)\hat{k} = 15\hat{i} - 21\hat{j} + 33\hat{k}$$.

$$\vec{A} \cdot (\vec{B} \times \vec{C}) = -15x + (-6)(-21) + (-2)(33) = -15x + 126 - 66 = -15x + 60 = 0$$.

$$x = 4$$.

The answer is 4.