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

Direction of line: $\vec{b}=\hat{i}+2\hat{j}+\lambda \hat{k}$.Normal to plane: $\vec{n}=\hat{i}+2\hat{j}+3\hat{k}$.The angle $\theta$ between line and plane is given by $\sin \theta =\frac{|\vec{b}\cdot \vec{n}|}{|\vec{b}||\vec{n}|}$.Given $\cos \theta =\sqrt{5/14}$, then $\sin \theta =\sqrt{1-5/14}=\sqrt{9/14}=3/\sqrt{14}$.
$$\frac{|1(1)+2(2)+\lambda (3)|}{\sqrt{1^2+2^2+{\lambda }^2}\sqrt{1^2+2^2+3^2}}=\frac{3}{\sqrt{14}}$$
$$\frac{|5+3\lambda |}{\sqrt{5+{\lambda }^2}\sqrt{14}}=\frac{3}{\sqrt{14}}$$
$|5+3\lambda |=3\sqrt{5+{\lambda }^2}$ Square both sides: $25+9{\lambda }^2+30\lambda =9(5+{\lambda }^2)$$25+9{\lambda }^2+30\lambda =45+9{\lambda }^2$$30\lambda =20⟹\lambda =2/3$.