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If the angle between the line $x=\frac{y-1}{2}=\frac{z-3}{\lambda}$ and the plane $x+2 y+3 z=4$ is $\cos ^{-1}\left(\sqrt{\frac{5}{14}}\right)$, then $\lambda$ equals
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$\frac{2}{3}$
$\frac{2}{3}$
$\cos \theta=\sqrt{\frac{5}{14}}$
$\sin \theta=\frac{3}{\sqrt{14}}$
$\sin \theta=\frac{1+4+3 \lambda}{\sqrt{1+4+\lambda^2} \sqrt{1+4+9}}$
$\frac{3}{\sqrt{14}}=\frac{5+3 \lambda}{\sqrt{5+\lambda^2} \sqrt{14}} \Rightarrow \lambda=\frac{2}{3}$
$\sin \theta=\frac{3}{\sqrt{14}}$
$\sin \theta=\frac{1+4+3 \lambda}{\sqrt{1+4+\lambda^2} \sqrt{1+4+9}}$
$\frac{3}{\sqrt{14}}=\frac{5+3 \lambda}{\sqrt{5+\lambda^2} \sqrt{14}} \Rightarrow \lambda=\frac{2}{3}$
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