Let $$\vec{a} = \alpha \hat{i} + 3\hat{j} - \hat{k}$$, $$\vec{b} = 3\hat{i} - \beta \hat{j} + 4\hat{k}$$ and $$\vec{c} = \hat{i} + 2\hat{j} - 2\hat{k}$$ where $$\alpha, \beta \in R$$. If the projection of $$\vec{a}$$ on $$\vec{c}$$ is $$\frac{10}{3}$$ and $$\vec{b} \times \vec{c} = -6\hat{i} + 10\hat{j} + 7\hat{k}$$, then the value of $$\alpha + \beta$$ equal to

We are given vectors $$\vec{a} = \alpha\hat{i} + 3\hat{j} - \hat{k}$$, $$\vec{b} = 3\hat{i} - \beta\hat{j} + 4\hat{k}$$, and $$\vec{c} = \hat{i} + 2\hat{j} - 2\hat{k}$$.

To find $$\alpha$$ using the projection condition, note that the projection of $$\vec{a}$$ on $$\vec{c}$$ is $$\frac{\vec{a} \cdot \vec{c}}{|\vec{c}|} = \frac{10}{3}$$. We have $$|\vec{c}| = \sqrt{1 + 4 + 4} = 3$$ and $$\vec{a} \cdot \vec{c} = \alpha + 6 + 2 = \alpha + 8$$, so $$\frac{\alpha + 8}{3} = \frac{10}{3}$$ implies $$\alpha + 8 = 10 \Rightarrow \alpha = 2$$.

Next, we determine $$\beta$$ using the cross product condition by computing $$\vec{b} \times \vec{c} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & -\beta & 4 \\ 1 & 2 & -2 \end{vmatrix}$$, which expands to $$\hat{i}(2\beta - 8) - \hat{j}(-6 - 4) + \hat{k}(6 + \beta) = (2\beta - 8)\hat{i} + 10\hat{j} + (6 + \beta)\hat{k}$$. Since $$\vec{b} \times \vec{c} = -6\hat{i} + 10\hat{j} + 7\hat{k}$$, equating components yields from the $$\hat{i}$$ component $$2\beta - 8 = -6 \Rightarrow \beta = 1$$ and from the $$\hat{k}$$ component $$6 + \beta = 7 \Rightarrow \beta = 1$$, confirming consistency.

Finally, $$\alpha + \beta = 2 + 1 = 3$$, so the answer is 3.