Let $\bar{a}=\alpha \hat{i}+3 \hat{\mathrm{j}}-\hat{\mathrm{k}}, \overline{\mathrm{b}}=3 \hat{i}-\hat{\mathrm{j}}+\beta \hat{\mathrm{k}}$ and $\overline{\mathrm{c}}=\hat{i}+2 \hat{\mathrm{j}}-2 \hat{\mathrm{k}}$ where $\alpha, \beta \in \mathbb{R}$, be three vectors. If the projection of $\bar{a}$ on $\bar{c}$ is $\frac{10}{3}$ and $\overline{\mathrm{b}} \times \overline{\mathrm{c}}=-6 \hat{i}+10 \hat{\mathrm{j}}+7 \hat{\mathrm{k}}$, then the value of $(\alpha+\beta)$ is equal to

Given vectors $\bar{a}=\alpha \hat{i}+3 \hat{j}-\hat{k}$, $\bar{b}=3 \hat{i}-\hat{j}+\beta \hat{k}$, and $\bar{c}=\hat{i}+2 \hat{j}-2 \hat{k}$.

The projection of $\bar{a}$ on $\bar{c}$ is $\frac{\bar{a} \cdot \bar{c}}{|\bar{c}|} = \frac{\alpha + 8}{3}$, since $\bar{a} \cdot \bar{c} = \alpha + 6 + 2$ and $|\bar{c}| = 3$.
Given this equals $\frac{10}{3}$, we find $\alpha = 2$.

Computing $\bar{b} \times \bar{c}$ yields $(2 - 2\beta)\hat{i} + (6 + \beta)\hat{j} + 7\hat{k}$.
Setting this equal to $-6\hat{i} + 10\hat{j} + 7\hat{k}$ and comparing coefficients gives $\beta = 4$.

Therefore, $\alpha + \beta = 2 + 4 = 6$.

$6$