Lines $\vec{r} = \vec{a} + \lambda\vec{b}$ and $\vec{r} = \vec{b} + \mu\vec{a}$ intersect at point $(2, 4, -4)$. If $|\vec{a} - \vec{b}| = 4$, then $\vec{a} \cdot \vec{b} =$

Let the point of intersection be $\vec{p} = 2\hat{i} + 4\hat{j} - 4\hat{k}$.

Since the lines intersect at $\vec{p}$, there exist scalars $\lambda$ and $\mu$ such that:

$\vec{a} + \lambda\vec{b} = \vec{b} + \mu\vec{a}$

$\Rightarrow (1 - \mu)\vec{a} + (\lambda - 1)\vec{b} = \vec{0}$

Assuming $\vec{a}$ and $\vec{b}$ are non-collinear, we get $1 - \mu = 0$ and $\lambda - 1 = 0$, which gives $\lambda = 1$ and $\mu = 1$.

Thus, the position vector of the point of intersection is:

$\vec{p} = \vec{a} + \vec{b}$

The magnitude squared of $\vec{p}$ is:

$|\vec{a} + \vec{b}|^2 = 2^2 + 4^2 + (-4)^2 = 4 + 16 + 16 = 36$

We are given $|\vec{a} - \vec{b}| = 4$, so:

$|\vec{a} - \vec{b}|^2 = 16$

Using the vector identity:

$|\vec{a} + \vec{b}|^2 - |\vec{a} - \vec{b}|^2 = 4\vec{a} \cdot \vec{b}$

Substituting the values:

$36 - 16 = 4\vec{a} \cdot \vec{b}$

$20 = 4\vec{a} \cdot \vec{b}$

$\vec{a} \cdot \vec{b} = 5$

Answer: $5$