Two adjacent sides of a parallelogram ABCD are given by
$$\overrightarrow{AB} = 2\hat{i} + 10\hat{j} + 11\hat{k}$$ and $$\overrightarrow{AD} = -\hat{i} + 2\hat{j} + 2\hat{k}$$
The side AD is rotated by an acute angle $$\alpha$$ in the plane of the parallelogram so that AD becomes AD'. If AD' makes a right angle with the side AB, then the cosine of the angle $$\alpha$$ is given by

A

$$\frac{8}{9}$$

B

$$\frac{\sqrt{17}}{9}$$

C

$$\frac{1}{9}$$

D

$$\frac{4\sqrt{5}}{9}$$