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(C) Let the direction cosines of the line be $(\cos 	heta, \cos \beta, \cos 	heta)$.Since the sum of the squares of the direction cosines is $1$,we

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$\frac{3}{5}$

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(C) Let the direction cosines of the line be $(\cos 	heta, \cos \beta, \cos 	heta)$.Since the sum of the squares of the direction cosines is $1$,we have:$\cos^2 	heta + \cos^2 \beta + \cos^2 	heta = 1$$2\cos^2 	heta + \cos^2 \beta = 1$Using the identity $\cos^2 \alpha = 1 - \sin^2 \alpha$,we get:$2(1 - \sin^2 	heta) + (1 - \sin^2 \beta) = 1$$3 - 2\sin^2 	heta - \sin^2 \beta = 1$Given $\sin^2 \beta = 3\sin^2 	heta$,substitute this into the equation:$3 - 2\sin^2 	heta - 3\sin^2 	heta = 1$$3 - 5\sin^2 	heta = 1$$5\sin^2 	heta = 2$$\sin^2 	heta = \frac{2}{5}$Now,calculate $\cos^2 	heta$:$\cos^2 	heta = 1 - \sin^2 	heta = 1 - \frac{2}{5} = \frac{3}{5}$

$A$ line makes the same angle $\theta$ with each of the $x$ and $z$ axes. If the angle $\beta$ which it makes with the $y$-axis is such that $\sin^2 \beta = 3\sin^2 \theta$,then $\cos^2 \theta$ is-

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