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(B) Let the direction cosines of the line $\overrightarrow{OR}$ be $l, m, n$. Since the line makes angles $	heta_{1}, 	heta_{2}, 	heta_{3}$ with th

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$2$

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(B) Let the direction cosines of the line $\overrightarrow{OR}$ be $l, m, n$. Since the line makes angles $	heta_{1}, 	heta_{2}, 	heta_{3}$ with the planes $XOY, YOZ, ZOX$ respectively,the angles with the normals to these planes (which are the $Z, X, Y$ axes) are $\frac{\pi}{2}-	heta_{1}, \frac{\pi}{2}-	heta_{2}, \frac{\pi}{2}-	heta_{3}$. Thus,$|l| = \cos(\frac{\pi}{2}-	heta_{2}) = \sin 	heta_{2}$,$|m| = \cos(\frac{\pi}{2}-	heta_{3}) = \sin 	heta_{3}$,and $|n| = \cos(\frac{\pi}{2}-	heta_{1}) = \sin 	heta_{1}$. We know that for direction cosines,$l^{2}+m^{2}+n^{2}=1$. Substituting the values,we get $\sin^{2} 	heta_{2} + \sin^{2} 	heta_{3} + \sin^{2} 	heta_{1} = 1$. Using the identity $\sin^{2} 	heta = 1 - \cos^{2} 	heta$,we have $(1-\cos^{2} 	heta_{2}) + (1-\cos^{2} 	heta_{3}) + (1-\cos^{2} 	heta_{1}) = 1$. $3 - (\cos^{2} 	heta_{1} + \cos^{2} 	heta_{2} + \cos^{2} 	heta_{3}) = 1$. Therefore,$\cos^{2} 	heta_{1} + \cos^{2} 	heta_{2} + \cos^{2} 	heta_{3} = 3 - 1 = 2$.

If the line $\overrightarrow{OR}$ makes angles $\theta_{1}, \theta_{2}, \theta_{3}$ with the planes $XOY, YOZ, ZOX$ respectively,then $\cos ^{2} \theta_{1}+\cos ^{2} \theta_{2}+\cos ^{2} \theta_{3}$ is equal to

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