If a ray makes the angles $\mathrm{\alpha },\mathrm{\beta },\mathrm{\gamma }\text{ and }\mathrm{\delta }$ with four diagonals of a cube then prove that ${\cos }^2\mathrm{\alpha }+{\cos }^2\mathrm{\beta }+{\cos }^2\mathrm{\gamma }+{\cos }^2\mathrm{\delta }=4/3$

Let ‘a’ be the length of each side of cube. Let one vertex of the cube is origin ‘O’ and the coordinate axes along the 3 edges $\overline{\mathrm{OA}},\overline{\mathrm{OB}}\text{ and }$$\overline{\mathrm{OC}}$ are passing through origin. The 4 diagonals are $\overline{\mathrm{OG}},\overline{\mathrm{AF}},\overline{\mathrm{CE}}\text{ and }\overline{\mathrm{BD}}$
 
$\begin{array}{l}\mathrm{O}=(0,0,0),\mathrm{A}(\mathrm{a},0,0),\mathrm{B}(0,\mathrm{a},0),\mathrm{C}(0,0,\mathrm{a})\\ \mathrm{D}=(\mathrm{a},0,\mathrm{a}),\mathrm{E}(\mathrm{a},\mathrm{a},0),\mathrm{F}(0,\mathrm{a},\mathrm{a}),\mathrm{G}(\mathrm{a},\mathrm{a},\mathrm{a})\end{array}$
$\text{ Let }(\mathrm{l},\mathrm{m},\mathrm{n})\text{ are d.c's of given ray }$
$\begin{array}{l}\mathrm{d}.\mathrm{r}’\mathrm{s} \mathrm{of} \overline{\mathrm{OG}}=\left({\mathrm{x}}_2-{\mathrm{x}}_1,{\mathrm{y}}_2-{\mathrm{y}}_1,{\mathrm{z}}_2-{\mathrm{z}}_1\right)\\ =(\mathrm{a}-0,\mathrm{a}-0,\mathrm{a}-0)=(\mathrm{a},\mathrm{a},\mathrm{a})\end{array}$
$\text{ dc's of }\overline{\mathrm{OG}}$
$\begin{array}{l}=\left(\frac{\mathrm{a}}{\sqrt{{\mathrm{a}}^2+{\mathrm{a}}^2+{\mathrm{a}}^2}},\frac{\mathrm{a}}{\sqrt{{\mathrm{a}}^2+{\mathrm{a}}^2+{\mathrm{a}}^2}},\frac{\mathrm{a}}{\sqrt{{\mathrm{a}}^2+{\mathrm{a}}^2+{\mathrm{a}}^2}}\right)\\ =\left(\frac{\mathrm{a}}{\sqrt{3}\mathrm{a}},\frac{\mathrm{a}}{\sqrt{3}\mathrm{a}},\frac{\mathrm{a}}{\sqrt{3}\mathrm{a}}\right)=\left(\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}}\right)\end{array}$
$\text{ dr's of }\overline{\mathrm{AF}}=(0-\mathrm{a},\mathrm{a}-0,\mathrm{a}-0)=(-\mathrm{a},\mathrm{a},\mathrm{a})$
$\text{ dc's of }\overline{\mathrm{AF}}$
$\begin{array}{l}=\left(\frac{-\mathrm{a}}{\sqrt{{\mathrm{a}}^2+{\mathrm{a}}^2+{\mathrm{a}}^2}},\frac{\mathrm{a}}{\sqrt{{\mathrm{a}}^2+{\mathrm{a}}^2+{\mathrm{a}}^2}},\frac{\mathrm{a}}{\sqrt{{\mathrm{a}}^2+{\mathrm{a}}^2+{\mathrm{a}}^2}}\right)\\ =\left(\frac{-\mathrm{a}}{\sqrt{3}\mathrm{a}},\frac{\mathrm{a}}{\sqrt{3}\mathrm{a}},\frac{\mathrm{a}}{\sqrt{3}\mathrm{a}}\right)=\left(-\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}}\right)\end{array}$
$\text{ dr's of }\overline{\mathrm{BD}}=(\mathrm{a}-0,0-\mathrm{a},\mathrm{a}-0)=(\mathrm{a},-\mathrm{a},\mathrm{a})$
$\text{ dc's of }\overline{\mathrm{BD}}$
$=\left(\frac{\mathrm{a}}{\sqrt{{\mathrm{a}}^2+{\mathrm{a}}^2+{\mathrm{a}}^2}},\frac{-\mathrm{a}}{\sqrt{{\mathrm{a}}^2+{\mathrm{a}}^2+{\mathrm{a}}^2}},\frac{\mathrm{a}}{\sqrt{{\mathrm{a}}^2+{\mathrm{a}}^2+{\mathrm{a}}^2}}\right)$
$=\left(\frac{\mathrm{a}}{\sqrt{3}\mathrm{a}},\frac{-\mathrm{a}}{\sqrt{3}\mathrm{a}},\frac{\mathrm{a}}{\sqrt{3}\mathrm{a}}\right)=\left(\frac{1}{\sqrt{3}},\frac{-1}{\sqrt{3}},\frac{1}{\sqrt{3}}\right)$
$\text{ dr's of }\overline{\mathrm{CE}}=(\mathrm{a}-0,\mathrm{a}-0,0-\mathrm{a})=(\mathrm{a},\mathrm{a},-\mathrm{a})$
$\text{ dc's of }\overline{\mathrm{CE}}$
$\begin{array}{l}=\left(\frac{\mathrm{a}}{\sqrt{{\mathrm{a}}^2+{\mathrm{a}}^2+{\mathrm{a}}^2}},\frac{\mathrm{a}}{\sqrt{{\mathrm{a}}^2+{\mathrm{a}}^2+{\mathrm{a}}^2}},\frac{-\mathrm{a}}{\sqrt{{\mathrm{a}}^2+{\mathrm{a}}^2+{\mathrm{a}}^2}}\right)\\ =\left(\frac{\mathrm{a}}{\sqrt{3}\mathrm{a}},\frac{\mathrm{a}}{\sqrt{3}\mathrm{a}},\frac{-\mathrm{a}}{\sqrt{3}\mathrm{a}}\right)=\left(\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{-1}{\sqrt{3}}\right)\end{array}$
$\text{ Let }\mathrm{\alpha },\mathrm{\beta },\mathrm{\gamma },\mathrm{\delta }\text{ are the angles made by the ray }$
$(\mathrm{l},\mathrm{m},\mathrm{n})\text{ with }\overline{\mathrm{OG}},\overline{\mathrm{AF}},\overline{\mathrm{BD}},\overline{\mathrm{CE}}\text{ respectively. Then }$
$\begin{array}{l}\cos \mathrm{\alpha }=\left|{\mathrm{l}}_1{\mathrm{l}}_2+{\mathrm{m}}_1{\mathrm{m}}_2+{\mathrm{n}}_1{\mathrm{n}}_2\right|\\ =\left|\mathrm{l}\left(\frac{1}{\sqrt{3}}\right)+\mathrm{m}\left(\frac{1}{\sqrt{3}}\right)+\mathrm{n}\left(\frac{1}{\sqrt{3}}\right)\right|=\left|\frac{\mathrm{l}+\mathrm{m}+\mathrm{n}}{\sqrt{3}}\right|\end{array}$
Similarly, we get $\cos \mathrm{\beta }=\left|\frac{-\mathrm{l}+\mathrm{m}+\mathrm{n}}{\sqrt{3}}\right|$
$\begin{array}{l}\cos \mathrm{\gamma }=\left|\frac{\mathrm{l}-\mathrm{m}+\mathrm{n}}{\sqrt{3}}\right|\text{ and }\cos \mathrm{\delta }=\left|\frac{\mathrm{l}+\mathrm{m}-\mathrm{n}}{\sqrt{3}}\right|\\ {\cos }^2\mathrm{\alpha }+{\cos }^2\mathrm{\beta }+{\cos }^2\mathrm{\gamma }+{\cos }^2\mathrm{\delta }=\\ \frac{(\mathrm{l}+\mathrm{m}+\mathrm{n}{)}^2}{3}+\frac{(-\mathrm{l}+\mathrm{m}+\mathrm{n}{)}^2}{3}+\frac{(\mathrm{l}-\mathrm{m}+\mathrm{n}{)}^2}{3}+\frac{(\mathrm{l}+\mathrm{m}-\mathrm{n}{)}^2}{3}\\ =\frac{4{\mathrm{l}}^2+4{\mathrm{m}}^2+4{\mathrm{n}}^2}{3}=\frac{4\left({\mathrm{l}}^2+{\mathrm{m}}^2+{\mathrm{n}}^2\right)}{3}\\ =\frac{4}{3}\left(\because {\mathrm{l}}^2+{\mathrm{m}}^2+{\mathrm{n}}^2=1\right)\end{array}$