A B C D E F G H is a cuboid with base A B C D . Let A(0,0,0),...

$\therefore$ Direction cosines of $A B=1,0,0$. Direction cosines of $\mathrm{AG}=$$\begin{aligned} & =\frac{12}{\sqrt{12^2+6^2+4^2}}, \frac{6}{\sqrt{12^2+6^2+4^2}}, \frac{4}{\sqrt{12^2+6^2+4^2}} \\ & =\frac{6}{7}, \frac{3}{7}, \frac{2}{7}\end{aligned}$$$Direction cosines of AC$$\begin{aligned} & =\frac{12}{\sqrt{12^2+6^2}}, \frac{6}{\sqrt{12^2+6^2}}, 0 \\ & \frac{2}{\sqrt{5}}, \frac{1}{\sqrt{5}}, 0 \end{aligned}$$\begin{aligned} & \therefore \cos \alpha=\frac{6}{7}+0+0=\frac{6}{7} \\ & \cos \beta=\frac{6}{7} \times \frac{2}{\sqrt{5}}+\frac{3}{7} \times \frac{1}{\sqrt{5}}+\frac{2}{7} \times 0=\frac{15}{7 \sqrt{5}} \end{aligned}$Now,$\cos 2 \alpha+\cos 2 \beta=2$\left[\cos ^2 \alpha+\cos ^2 \beta\right]-2=2$\left[\frac{36}{49}+\frac{225}{49 \times 5}\right]-2=\frac{64}{49}$