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

$\frac{2}{3}$

Let θ be the angle between the line $\frac{x-0}{1}=\frac{y-1}{2}=\frac{z-3}{λ}$ and the plane $x + 2y + 3z - 4 = 0.$ Then,

$cos(\frac{\pi}{2}- \theta )=\frac{1×1+2×2+3 ×λ}{\sqrt{1+4+9}\sqrt{1+4+λ^2}} \left[∵ cos (\frac{\pi}{2}\theta ) = \frac{\vec{b}.\vec{n}}{|\vec{b}||\vec{n}|} \right]$

$⇒ sin \theta = \frac{5 + 3 λ}{\sqrt{14}\sqrt{5+λ^2}}$

$⇒ sin^2 \theta = \frac{(5 + 3λ)^2}{14(5+λ^2)}$

$⇒ 1- \frac{15}{14}= \frac{(5 + 3λ)^2}{14(5+λ^2)}$       $\left[cos \theta = \sqrt{\frac{5}{14}}(given)\right]$

$⇒ 45 + 9λ^2 = 25 + 30 λ + 9 λ^2⇒ λ = \frac{2}{3}$