Vector Algebra

Let $\vec{a}$ and $\vec{b}$ be linearly independent vectors such that
$|\vec{a}| = \sqrt{3}, |\vec{b}| = 3$ and $|\vec{a} - \vec{b}| = 4$.
If $\vec{a} \times (2\hat{i} + 2\hat{j} - \hat{k}) = (2\hat{i} + 2\hat{j} - \hat{k}) \times \vec{b}$ and $|(\vec{a} + \vec{b}) \cdot (3\hat{i} + 4\hat{j} + 2\hat{k})| = \sqrt{\lambda}$, then $\lambda = \ldots$