Vector and Algebra

Since $a, b, c$ are direction cosines of a line, so $a^2 + b^2 + c^2 = 1$

We know that $(a + b + c)^2 \ge 0$

$\implies$ $a^2 + b^2 + c^2 + 2(ab + bc + ca) \ge 0$

$\implies$ $ab+bc+ca \ge \frac{-1}{2}$

Now to calculate angle between $\vec{A}$ & $\vec{B}$ we should use dot product of two vectors.

$\implies$ $\vec{A}.\vec{B} = |\vec{A}||\vec{B}|cos\theta$

$\implies$ $ab+bc+ca = (\sqrt{a^2 + b^2 + c^2})(\sqrt{a^2 + b^2 + c^2})cos\theta$

$\implies$ $cos\theta \ge \frac{-1}{2}$

Hence maximum possible value of $\theta \in [0,\pi]$ is $\boxed{120^o}$