If the components of $\overrightarrow{\mathbf{a}}=\alpha \hat{i}+\beta \hat{j}+\gamma \hat{k}$ along and perpendicular to $\overrightarrow{\mathbf{b}}=3 \hat{i}+\hat{j}-\hat{k}$ 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 equal to: