Let \(\quad \overline{\mathrm{a}}=\alpha \hat{\mathrm{i}}+3 \hat{\mathrm{j}}-\hat{\mathrm{k}}, \quad \overline{\mathrm{b}}=3 \hat{\mathrm{i}}-\beta \hat{\mathrm{j}}+4 \hat{\mathrm{k}} \quad\) and \(\overline{\mathrm{c}}=\hat{\mathrm{i}}+2 \hat{\mathrm{j}}-2 \hat{\mathrm{k}}\), where \(\alpha, \beta \in \mathbb{R}\), be three vectors. If the projection at \(\overline{\mathrm{a}}\) on \(\overline{\mathrm{c}}\) is \(\frac{10}{3}\) and \(\overline{\mathrm{b}} \times \overline{\mathrm{c}}=-6 \hat{\mathrm{i}}+10 \hat{\mathrm{j}}+7 \hat{\mathrm{k}}\), then the value of \(\alpha^2+\beta^2-\alpha \beta\) is equal to

Projection of \(\overline{\mathrm{a}}\) on \(\overline{\mathrm{c}}\) is \(\frac{10}{3}\)
\(\begin{aligned}
& \frac{\overline{\mathrm{a}} \cdot \overline{\mathrm{c}}}{|\overline{\mathrm{c}}|}=\frac{10}{3} \\
& \Rightarrow \frac{(\alpha \hat{\mathrm{i}}+3 \hat{\mathrm{j}}-\hat{\mathrm{k}}) \cdot(\hat{\mathrm{i}}+2 \hat{\mathrm{j}}-2 \hat{\mathrm{k}})}{\sqrt{1+4+4}}=\frac{10}{3} \\
& \Rightarrow \frac{\alpha+3(2)-1(-2)}{\sqrt{9}}=\frac{10}{3} \\
& \Rightarrow \frac{\alpha+8}{3}=\frac{10}{3} \\
& \Rightarrow \alpha=2 \\
& \overline{\mathrm{~b}} \times \overline{\mathrm{c}}=-6 \hat{\mathrm{i}}+10 \hat{\mathrm{j}}+7 \hat{\mathrm{k}} \\
& \Rightarrow\left|\begin{array}{cc}
\hat{\mathrm{i}} & \hat{\mathrm{j}} \quad \hat{\mathrm{k}} \\
3 & -\beta \\
1 & \quad 2
\end{array} \quad-2\right|=-6 \hat{\mathrm{i}}+10 \hat{\mathrm{j}}+7 \hat{\mathrm{k}} \\
& \Rightarrow(2 \beta-8) \hat{\mathrm{i}}+10 \hat{\mathrm{j}}+(6+\beta) \hat{\mathrm{k}}=-6 \hat{\mathrm{i}}+10 \hat{\mathrm{j}}+7 \hat{\mathrm{k}} \\
& \Rightarrow 6+\beta=7 \Rightarrow \beta=1 \\
& \alpha^2+\beta^2-\alpha \beta=4+1-2=3
\end{aligned}\)