The line passing through the points (1, 1, -1 | vedclass

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(C) The direction vector of the line passing through $P(1, 1, -1)$ and $Q(3, -1, 0)$ is $\vec{v} = (3-1)\hat{i} + (-1-1)\hat{j} + (0-(-1))\hat{k} = 2\

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$15$

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(C) The direction vector of the line passing through $P(1, 1, -1)$ and $Q(3, -1, 0)$ is $\vec{v} = (3-1)\hat{i} + (-1-1)\hat{j} + (0-(-1))\hat{k} = 2\hat{i} - 2\hat{j} + \hat{k}$.The normal vector to the plane $\sqrt{\lambda} x + 3y + 6z = 17$ is $\vec{n} = \sqrt{\lambda}\hat{i} + 3\hat{j} + 6\hat{k}$.The angle $	heta$ between a line with direction $\vec{v}$ and a plane with normal $\vec{n}$ is given by $\sin 	heta = \frac{|\vec{v} \cdot \vec{n}|}{|\vec{v}| |\vec{n}|}$.Given $	heta = \operatorname{Tan}^{-1}\left(\frac{1}{\sqrt{8}}ight)$,we have $	an 	heta = \frac{1}{\sqrt{8}}$,which implies $\sin 	heta = \frac{1}{\sqrt{1^2 + (\sqrt{8})^2}} = \frac{1}{3}$.Calculating the dot product: $\vec{v} \cdot \vec{n} = (2)(\sqrt{\lambda}) + (-2)(3) + (1)(6) = 2\sqrt{\lambda} - 6 + 6 = 2\sqrt{\lambda}$.Calculating magnitudes: $|\vec{v}| = \sqrt{2^2 + (-2)^2 + 1^2} = \sqrt{4+4+1} = 3$ and $|\vec{n}| = \sqrt{(\sqrt{\lambda})^2 + 3^2 + 6^2} = \sqrt{\lambda + 9 + 36} = \sqrt{\lambda + 45}$.Substituting into the formula: $\frac{1}{3} = \frac{|2\sqrt{\lambda}|}{3 \sqrt{\lambda + 45}}$.$1 = \frac{2\sqrt{\lambda}}{\sqrt{\lambda + 45}} \implies \sqrt{\lambda + 45} = 2\sqrt{\lambda}$.Squaring both sides: $\lambda + 45 = 4\lambda \implies 3\lambda = 45 \implies \lambda = 15$.

The line passing through the points $(1, 1, -1)$ and $(3, -1, 0)$ makes an angle of $\operatorname{Tan}^{-1}\left(\frac{1}{\sqrt{8}}\right)$ with the plane $\sqrt{\lambda} x + 3y + 6z = 17$. Then $\lambda =$

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