The cosine of the angle of the triangle with vertices A(1, -1, 2), B(6, 11, 2) and C(1, 2, 6), is

$\frac{63}{65}$ $\frac{36}{65}$ $\frac{16}{65}$ $\frac{13}{65}$

$\frac{36}{65}$

We have, $c = AB = 13, b = AC = 5 $ and $ a = BC = \sqrt{122}$ $∴ cos A = \frac{b^2+c^2-a^2}{2bc}⇒ cos A = \frac{25+169-122}{2×5×13}=\frac{36}{65}$