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 :

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Given that the vector $\mathbf{a}$ can be resolved into a component along vector $$\mathbf{b}$$ and a component perpendicular to vector $$\mathbf{b}$$:

$$\mathbf{a} = \mathbf{a}_{\parallel} + \mathbf{a}_{\perp}$$

$$\mathbf{a}$$ = $$\frac{16}{11}(3\hat{i} + \hat{j} - \hat{k})$$ + $$\frac{1}{11}(-4\hat{i} - 5\hat{j} - 17\hat{k})$$

Taking $$\frac{1}{11}$$ as a common factor:

$$\mathbf{a} = \frac{1}{11} \left[ (48\hat{i} + 16\hat{j} - 16\hat{k}) + (-4\hat{i} - 5\hat{j} - 17\hat{k}) \right]$$

$$\mathbf{a} = \frac{1}{11} (44\hat{i} + 11\hat{j} - 33\hat{k})$$ $$\mathbf{a} = 4\hat{i} + \hat{j} - 3\hat{k}$$

Let $$\mathbf{a} = \alpha\hat{i} + \beta\hat{j} + \gamma\hat{k}$$. By comparing coefficients, we get:

• $$\alpha = 4$$
• $$\beta = 1$$
• $$\gamma = -3$$

Now, calculate $$\alpha^2 + \beta^2 + \gamma^2$$:

$$\alpha^2 + \beta^2 + \gamma^2 = (4)^2 + (1)^2 + (-3)^2$$ $$\alpha^2 + \beta^2 + \gamma^2 = 16 + 1 + 9 = 26$$