If the angle theta between the line frac{x + | vedclass

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(C) The angle $	heta$ between a line with direction vector $\vec{b} = (1, 2, 2)$ and a plane with normal vector $\vec{n} = (2, -1, \sqrt{\lambda})$ i

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$\frac{5}{3}$

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(C) The angle $	heta$ between a line with direction vector $\vec{b} = (1, 2, 2)$ and a plane with normal vector $\vec{n} = (2, -1, \sqrt{\lambda})$ is given by $\sin 	heta = \frac{|\vec{b} \cdot \vec{n}|}{|\vec{b}| |\vec{n}|}$.Given $\sin 	heta = \frac{1}{3}$,we have $\frac{1}{3} = \frac{|(1)(2) + (2)(-1) + (2)(\sqrt{\lambda})|}{\sqrt{1^2 + 2^2 + 2^2} \sqrt{2^2 + (-1)^2 + (\sqrt{\lambda})^2}}$.Simplifying the expression: $\frac{1}{3} = \frac{|2 - 2 + 2\sqrt{\lambda}|}{\sqrt{9} \sqrt{5 + \lambda}}$.$\frac{1}{3} = \frac{2\sqrt{\lambda}}{3\sqrt{5 + \lambda}}$.Squaring both sides: $\frac{1}{9} = \frac{4\lambda}{9(5 + \lambda)}$.$5 + \lambda = 4\lambda$.$3\lambda = 5 \Rightarrow \lambda = \frac{5}{3}$.

If the angle $\theta$ between the line $\frac{x + 1}{1} = \frac{y - 1}{2} = \frac{z - 2}{2}$ and the plane $2x - y + \sqrt{\lambda} z + 4 = 0$ is such that $\sin \theta = \frac{1}{3}$,the value of $\lambda$ is

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