If the components of $$\overrightarrow{a} = \alpha \hat{i}+\beta \hat{j}+\gamma \hat{k}$$ along and perpendicular to $$\overrightarrow{b}= 3\hat{i}+\hat{j}-\hat{k}$$ respectively, are $$frac{16}{11}(3\hat{i}+\hat{j}-\hat{k})$$ and $$frac{1}{11}(-4\hat{i}-5\hat{j}-17\hat{k})$$, then $$\alpha^{2} + \beta^{2} + \gamma^{2}$$ is equals to :

A

26

B

18

C

23

D

16