Let bar{a}=alpha hat{i}+3 hat{j}-hat{k},bar{b | vedclass

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(C) The projection of $\overline{a}$ on $\overline{c}$ is given by $\frac{\overline{a} \cdot \overline{c}}{|\overline{c}|}$.Given $\overline{a} = \alp

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$5$

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(C) The projection of $\overline{a}$ on $\overline{c}$ is given by $\frac{\overline{a} \cdot \overline{c}}{|\overline{c}|}$.Given $\overline{a} = \alpha \hat{i} + 3 \hat{j} - \hat{k}$ and $\overline{c} = \hat{i} + 2 \hat{j} - 2 \hat{k}$.$\overline{a} \cdot \overline{c} = (\alpha)(1) + (3)(2) + (-1)(-2) = \alpha + 6 + 2 = \alpha + 8$.$|\overline{c}| = \sqrt{1^2 + 2^2 + (-2)^2} = \sqrt{1 + 4 + 4} = \sqrt{9} = 3$.So,$\frac{\alpha + 8}{3} = \frac{10}{3} \implies \alpha + 8 = 10 \implies \alpha = 2$.Now,$\overline{b} 	imes \overline{c} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 3 & -\beta & 4 \ 1 & 2 & -2 \end{vmatrix} = \hat{i}(2\beta - 8) - \hat{j}(-6 - 4) + \hat{k}(6 + \beta) = (2\beta - 8)\hat{i} + 10\hat{j} + (6 + \beta)\hat{k}$.Comparing this with $-6\hat{i} + 10\hat{j} + 7\hat{k}$,we get $2\beta - 8 = -6 \implies 2\beta = 2 \implies \beta = 1$.Finally,$2\alpha + \beta = 2(2) + 1 = 4 + 1 = 5$.

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