Let $\vec{a}=\hat{i}-\hat{j}+\hat{k}$, $\vec{b}=\hat{i}-2\hat{j}-2\hat{k}$, $\vec{c}=6\hat{i}+3\hat{j}-2\hat{k}$ be three vectors. If $\vec{d}$ is a vector perpendicular to both $\vec{a}$, $\vec{b}$ and $|\vec{d}\times\vec{c}|=14$, then $|\vec{d}\cdot\vec{c}|=$
Show Hint
A vector perpendicular to two given vectors $\vec{a}$ and $\vec{b}$ is always of the form $\lambda(\vec{a} \times \vec{b})$. The scalar triple product $(\vec{a} \times \vec{b}) \cdot \vec{c}$ gives the volume of the parallelepiped formed by the three vectors.