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(D) The line $L$ is given by $\frac{x}{1}=\frac{y}{0}=\frac{z}{-1} = \lambda$. So,any point on $L$ is $N(\lambda, 0, -\lambda)$.Since $PN \perp L$,the

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$\frac{1}{\sqrt{3}}$

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(D) The line $L$ is given by $\frac{x}{1}=\frac{y}{0}=\frac{z}{-1} = \lambda$. So,any point on $L$ is $N(\lambda, 0, -\lambda)$.Since $PN \perp L$,the vector $\vec{PN} = (\lambda-1, -2, -\lambda+1)$ is perpendicular to the direction vector of $L$,which is $\vec{v} = (1, 0, -1)$.Thus,$(\lambda-1)(1) + (-2)(0) + (-\lambda+1)(-1) = 0 \Rightarrow \lambda-1 + \lambda-1 = 0 \Rightarrow 2\lambda = 2 \Rightarrow \lambda = 1$.So,$N = (1, 0, -1)$ and $\vec{PN} = (0, -2, 0)$.Now,let the line through $P$ parallel to the plane $x+y+2z=0$ meet $L$ at $Q(\mu, 0, -\mu)$.The vector $\vec{PQ} = (\mu-1, -2, -\mu+1)$. Since this line is parallel to the plane,$\vec{PQ}$ is perpendicular to the normal of the plane $\vec{n} = (1, 1, 2)$.Thus,$(\mu-1)(1) + (-2)(1) + (-\mu+1)(2) = 0 \Rightarrow \mu-1 - 2 - 2\mu + 2 = 0 \Rightarrow -\mu - 1 = 0 \Rightarrow \mu = -1$.So,$Q = (-1, 0, 1)$ and $\vec{PQ} = (-2, -2, 2)$.The angle $\alpha$ between $\vec{PN} = (0, -2, 0)$ and $\vec{PQ} = (-2, -2, 2)$ is given by $\cos \alpha = \frac{|\vec{PN} \cdot \vec{PQ}|}{|\vec{PN}| |\vec{PQ}|}$.$|\vec{PN}| = \sqrt{0^2 + (-2)^2 + 0^2} = 2$.$|\vec{PQ}| = \sqrt{(-2)^2 + (-2)^2 + 2^2} = \sqrt{4+4+4} = \sqrt{12} = 2\sqrt{3}$.$\vec{PN} \cdot \vec{PQ} = (0)(-2) + (-2)(-2) + (0)(2) = 4$.$\cos \alpha = \frac{4}{2 	imes 2\sqrt{3}} = \frac{4}{4\sqrt{3}} = \frac{1}{\sqrt{3}}$.

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