Let the line ell: x = frac{1-y}{-2} = frac{z- | vedclass

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(A) The line $\ell$ is given by $x = \frac{y-1}{2} = \frac{z-3}{\lambda}$.The direction ratios of the line $\ell$ are $(1, 2, \lambda)$.The normal vec

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$11$

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(A) The line $\ell$ is given by $x = \frac{y-1}{2} = \frac{z-3}{\lambda}$.The direction ratios of the line $\ell$ are $(1, 2, \lambda)$.The normal vector of the plane $P: x + 2y + 3z = 4$ is $\vec{n} = (1, 2, 3)$.The angle $	heta$ between the line and the plane is given by $\sin 	heta = \frac{|\vec{v} \cdot \vec{n}|}{|\vec{v}| |\vec{n}|}$,where $\vec{v} = (1, 2, \lambda)$.Given $\cos 	heta = \sqrt{\frac{5}{14}}$,we have $\sin 	heta = \sqrt{1 - \cos^2 	heta} = \sqrt{1 - \frac{5}{14}} = \sqrt{\frac{9}{14}} = \frac{3}{\sqrt{14}}$.Thus,$\frac{|1(1) + 2(2) + 3(\lambda)|}{\sqrt{1^2 + 2^2 + \lambda^2} \sqrt{1^2 + 2^2 + 3^2}} = \frac{3}{\sqrt{14}}$.$\frac{|5 + 3\lambda|}{\sqrt{5 + \lambda^2} \sqrt{14}} = \frac{3}{\sqrt{14}} \Rightarrow |5 + 3\lambda| = 3\sqrt{5 + \lambda^2}$.Squaring both sides: $(5 + 3\lambda)^2 = 9(5 + \lambda^2) \Rightarrow 25 + 30\lambda + 9\lambda^2 = 45 + 9\lambda^2 \Rightarrow 30\lambda = 20 \Rightarrow \lambda = \frac{2}{3}$.Any point on the line $\ell$ is $(t, 2t + 1, \frac{2}{3}t + 3)$. Since it lies on the plane $x + 2y + 3z = 4$:$t + 2(2t + 1) + 3(\frac{2}{3}t + 3) = 4 \Rightarrow t + 4t + 2 + 2t + 9 = 4 \Rightarrow 7t = -7 \Rightarrow t = -1$.The point $(\alpha, \beta, \gamma)$ is $(-1, 2(-1) + 1, \frac{2}{3}(-1) + 3) = (-1, -1, \frac{7}{3})$.Therefore,$\alpha + 2\beta + 6\gamma = -1 + 2(-1) + 6(\frac{7}{3}) = -1 - 2 + 14 = 11$.

Let the line $\ell: x = \frac{1-y}{-2} = \frac{z-3}{\lambda}, \lambda \in R$ meet the plane $P: x + 2y + 3z = 4$ at the point $(\alpha, \beta, \gamma)$. If the angle between the line $\ell$ and the plane $P$ is $\cos^{-1}\left(\sqrt{\frac{5}{14}}\right)$,then $\alpha + 2\beta + 6\gamma$ is equal to

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