A line makes angles α, β, γ, δ with the four diagonals of a cube. Then $cos^2α+cos^2β+cos^2γ+cos^2δ$ is equal to

4/3

The direction ratios of the diagonal $\vec{OR}$ are (1, 1, 1).

⇒ direction cosine are $(\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}})$.

Similarly direction cosine of $\vec{AS}$ are $(\frac{1}{\sqrt{3}},-\frac{1}{\sqrt{3}},-\frac{1}{\sqrt{3}})$ and those of $\vec{BP}$ and $\vec{CQ}$ are $(\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},-\frac{1}{\sqrt{3}})$, $(\frac{1}{\sqrt{3}},-\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}})$.

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

$cosα=\frac{l+m+n}{\sqrt{3}},cosβ=\frac{l-m-n}{\sqrt{3}},cosγ=\frac{l+m-n}{\sqrt{3}},cosδ=\frac{l-m+n}{\sqrt{3}}$

$⇒cos^2α+cos^2β+cos^2γ+cos^2δ=\frac{4(l^2+m^2+n^2)}{3}=\frac{3}{4}$ (since $l^2+m^2+n^2= 1$).

Hence (B) is the correct answer.