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(B) Let the side of the cube be $a$. The coordinates of the vertices are $(0,0,0), (a,0,0), (0,a,0), (0,0,a), (a,a,0), (a,0,a), (0,a,a), (a,a,a)$.The

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$\frac{4}{3}$

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(B) Let the side of the cube be $a$. The coordinates of the vertices are $(0,0,0), (a,0,0), (0,a,0), (0,0,a), (a,a,0), (a,0,a), (0,a,a), (a,a,a)$.The four diagonals of the cube are the lines joining the opposite vertices. Let these be $d_1, d_2, d_3, d_4$.The direction ratios of these diagonals are $(1,1,1), (1,1,-1), (1,-1,1), (-1,1,1)$.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 $	heta$ between a line with direction cosines $(l, m, n)$ and a line with direction ratios $(a_1, b_1, c_1)$ is given by $\cos 	heta = \frac{|l a_1 + m b_1 + n c_1|}{\sqrt{a_1^2 + b_1^2 + c_1^2}}$.For the four diagonals,the angles $\alpha, \beta, \gamma, \delta$ satisfy:$\cos \alpha = \frac{l+m+n}{\sqrt{3}}$,$\cos \beta = \frac{l+m-n}{\sqrt{3}}$,$\cos \gamma = \frac{l-m+n}{\sqrt{3}}$,$\cos \delta = \frac{-l+m+n}{\sqrt{3}}$.Squaring and adding these,we get:$\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 $\frac{4}{3} 	imes 1 = \frac{4}{3}$.

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

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