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(B) Let the direction cosines of the line be $(l, m, n)$,where $l^2 + m^2 + n^2 = 1$. The four diagonals of a cube are along the vectors $(\pm 1, \pm

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$\frac{8}{3}$

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(B) Let the direction cosines of the line be $(l, m, n)$,where $l^2 + m^2 + n^2 = 1$. The four diagonals of a cube are along the vectors $(\pm 1, \pm 1, \pm 1)$. The unit vectors along the four diagonals are $\vec{d_1} = \frac{1}{\sqrt{3}}(1, 1, 1)$,$\vec{d_2} = \frac{1}{\sqrt{3}}(-1, 1, 1)$,$\vec{d_3} = \frac{1}{\sqrt{3}}(1, -1, 1)$,and $\vec{d_4} = \frac{1}{\sqrt{3}}(1, 1, -1)$. The cosine of the angle $	heta$ between the line and a diagonal vector $\vec{d}$ is given by $|l \cdot d_x + m \cdot d_y + n \cdot d_z|$. 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 these,we get $\cos^2 \alpha = \frac{1}{3}(l+m+n)^2$,$\cos^2 \beta = \frac{1}{3}(-l+m+n)^2$,$\cos^2 \gamma = \frac{1}{3}(l-m+n)^2$,and $\cos^2 \delta = \frac{1}{3}(l+m-n)^2$. Summing these: $\sum \cos^2 \alpha = \frac{1}{3} [ (l+m+n)^2 + (-l+m+n)^2 + (l-m+n)^2 + (l+m-n)^2 ]$. Expanding the squares: $\sum \cos^2 \alpha = \frac{1}{3} [ 4(l^2+m^2+n^2) ] = \frac{4}{3}(1) = \frac{4}{3}$. Using $\sin^2 	heta = 1 - \cos^2 	heta$,we have $\sum \sin^2 \alpha = 4 - \sum \cos^2 \alpha = 4 - \frac{4}{3} = \frac{8}{3}$.

If a line makes angles $\alpha, \beta, \gamma$ and $\delta$ with the four diagonals of a cube,then the value of $\sin^2 \alpha + \sin^2 \beta + \sin^2 \gamma + \sin^2 \delta$ is

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