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

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(D) The projection of $\bar{a}$ on $\bar{c}$ is given by $\frac{\bar{a} \cdot \bar{c}}{|\bar{c}|} = \frac{10}{3}$.Given $\bar{a} = \alpha \hat{i} + 3

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$6$

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

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