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

<p>$$\begin{aligned} & \text { let } \\ & \overrightarrow{\mathrm{a}}_{11}=\text { component of } \overrightarrow{\mathrm{a}} \text { along } \overrightarrow{\mathrm{b}} \\ & \overrightarrow{\mathrm{a}}_1=\text { component of } \overrightarrow{\mathrm{a}} \text { perpendicular to } \overrightarrow{\mathrm{b}} \\ & \overrightarrow{\mathrm{a}}_{11}=\frac{16}{11}(3 \hat{\mathrm{i}}+\hat{\mathrm{j}}-\hat{\mathrm{k}}) \\ & \overrightarrow{\mathrm{a}}_1=\frac{1}{11}(-4 \hat{\mathrm{i}}-5 \hat{\mathrm{j}}-17 \hat{\mathrm{k}}) \\ & \because \overrightarrow{\mathrm{a}}=\overrightarrow{\mathrm{a}}_{11}+\overrightarrow{\mathrm{a}}_1 \\ & \therefore \overrightarrow{\mathrm{a}}=\frac{16}{11}(3 \hat{\mathrm{i}}+\hat{\mathrm{j}}-\hat{\mathrm{k}})+\frac{1}{11}(-4 \hat{\mathrm{i}}-5 \hat{\mathrm{j}}-17 \hat{\mathrm{k}}) \\ & \quad=\frac{44}{11} \hat{\mathrm{i}}+\frac{11}{11} \hat{\mathrm{j}}-\frac{33}{11} \hat{\mathrm{k}} \\ & \overrightarrow{\mathrm{a}}=4 \hat{\mathrm{i}}+\hat{\mathrm{j}}-3 \hat{\mathrm{k}} \quad \\ & \begin{array}{l} \alpha=4 \qquad \beta=1\qquad \gamma=-3\\ \alpha^2+\beta^2+\gamma^2=16+1+9=26 \end{array} \end{aligned}$$</p>