A line makes angles $\alpha, \beta, \gamma, \delta$ with the four diagonals of a cube then $\cos ^2 \alpha+\cos ^2 \beta+\cos ^2 \gamma+\cos ^2 \delta=$

4/3

The direction ratios of the diagonal $\overrightarrow{OR}~ (1,1,1)$

Direction cosine are $\left(\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}\right)$

Similarly direction cosine of $\overrightarrow{AS}$ are $\left(\frac{1}{\sqrt{3}},-\frac{1}{\sqrt{3}},-\frac{1}{\sqrt{3}}\right)$

$\overrightarrow{BP}$ are $\left(\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}},-\frac{1}{\sqrt{3}}\right)$

$\overrightarrow{CQ}$ are $\left(\frac{1}{\sqrt{3}},-\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}\right)$

Let l, m, n be direction cosines of the line

$\cos \alpha=\frac{1+m+n}{\sqrt{3}}, \cos \beta=\frac{l-m-n}{\sqrt{3}}, \cos \gamma=\frac{l+m-n}{\sqrt{3}}, \cos \delta=\frac{l-m+n}{\sqrt{3}}$

$\cos ^2 \alpha+\cos ^2 \beta+\cos ^2 \gamma+\cos ^2 \delta=\frac{4\left(l^2+m^2+n^2\right)}{3}=\frac{4}{3}$     (since $l^2+m^2+n^2=1$)

Hence (2) is the correct answer.