$A$ line makes angles $\alpha, \beta, \gamma$ and $\delta$ with the diagonals of a cube. Prove that $\cos^{2} \alpha + \cos^{2} \beta + \cos^{2} \gamma + \cos^{2} \delta = \frac{4}{3}$.

(N/A) cube is a rectangular parallelepiped having equal length,breadth,and height.
Let the vertices of the cube be defined such that the diagonals are $OE, AF, BG,$ and $CD$.
Let the side length of the cube be $a$.
The coordinates of the vertices are $O(0,0,0), A(a,0,0), B(0,a,0), C(0,0,a), D(a,a,0), E(a,a,a), F(0,a,a), G(a,0,a)$.
The direction vectors of the four diagonals are:
$d_1 = (a, a, a) \implies \text{unit vector } \hat{d}_1 = \frac{1}{\sqrt{3}}(1, 1, 1)$
$d_2 = (-a, a, a) \implies \text{unit vector } \hat{d}_2 = \frac{1}{\sqrt{3}}(-1, 1, 1)$
$d_3 = (a, -a, a) \implies \text{unit vector } \hat{d}_3 = \frac{1}{\sqrt{3}}(1, -1, 1)$
$d_4 = (a, a, -a) \implies \text{unit vector } \hat{d}_4 = \frac{1}{\sqrt{3}}(1, 1, -1)$
Let the direction cosines of the given line be $(l, m, n)$,where $l^2 + m^2 + n^2 = 1$.
The cosine of the angle between the line and a diagonal with unit vector $\hat{d}$ is given by the dot product: $\cos \theta = |l \cdot d_x + m \cdot d_y + n \cdot d_z|$.
Thus,$\cos \alpha = \frac{1}{\sqrt{3}}(l+m+n)$,$\cos \beta = \frac{1}{\sqrt{3}}(-l+m+n)$,$\cos \gamma = \frac{1}{\sqrt{3}}(l-m+n)$,and $\cos \delta = \frac{1}{\sqrt{3}}(l+m-n)$.
Squaring and adding these values:
$\cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma + \cos^2 \delta = \frac{1}{3} [(l+m+n)^2 + (-l+m+n)^2 + (l-m+n)^2 + (l+m-n)^2]$
$= \frac{1}{3} [ (l^2+m^2+n^2 + 2lm + 2mn + 2nl) + (l^2+m^2+n^2 + 2lm - 2mn - 2nl) + (l^2+m^2+n^2 - 2lm - 2mn + 2nl) + (l^2+m^2+n^2 - 2lm + 2mn - 2nl) ]$
$= \frac{1}{3} [ 4(l^2+m^2+n^2) ]$
Since $l^2+m^2+n^2 = 1$,the sum is $\frac{4}{3}(1) = \frac{4}{3}$.