Let $ \vec{a} = \hat{i} + 2\hat{j} + \hat{k} $, $ \vec{b} = 3\hat{i} - 3\hat{j} + 3\hat{k} $, $ \vec{c} = 2\hat{i} - \hat{j} + 2\hat{k} $ and $ \vec{d} $ be a vector such that $ \vec{b} \times \vec{d} = \vec{c} \times \vec{d} $ and $ \vec{a} \cdot \vec{d} = 4 $. Then $ |\vec{a} \times \vec{d}|^2 $ is equal to _______
Consider two vectors $\vec{u} = 3\hat{i} - \hat{j}$ and $\vec{v} = 2\hat{i} + \hat{j} - \lambda \hat{k}$, $\lambda>0$. The angle between them is given by $\cos^{-1} \left( \frac{\sqrt{5}}{2\sqrt{7}} \right)$. Let $\vec{v} = \vec{v}_1 + \vec{v}_2$, where $\vec{v}_1$ is parallel to $\vec{u}$ and $\vec{v}_2$ is perpendicular to $\vec{u}$. Then the value $|\vec{v}_1|^2 + |\vec{v}_2|^2$ is equal to