The angle between the line $x=\frac{\mathrm{y}-1}{2}=\frac{\mathrm{z}-3}{\lambda}$ and the plane $x+2 y+3 z=6$ is $\cos ^{-1} \sqrt{\frac{5}{14}}$, then the value of $\lambda$ is

Direction vector of the line: $\vec{b} = \langle 1, 2, \lambda \rangle$

Normal vector of the plane: $\vec{n} = \langle 1, 2, 3 \rangle$

The angle $\theta$ between the line and plane satisfies $\sin \theta = \frac{|\vec{b} \cdot \vec{n}|}{\|\vec{b}\| \cdot \|\vec{n}\|}$

Given $\theta = \cos^{-1} \sqrt{\frac{5}{14}}$, then $\sin \theta = \sqrt{1 - \frac{5}{14}} = \frac{3}{\sqrt{14}}$

Compute $\vec{b} \cdot \vec{n} = 1 + 4 + 3\lambda = 5 + 3\lambda$

Magnitudes: $\|\vec{b}\| = \sqrt{5 + \lambda^2}$, $\|\vec{n}\| = \sqrt{14}$

Substitute into the sine formula:
$\frac{3}{\sqrt{14}} = \frac{|5 + 3\lambda|}{\sqrt{5 + \lambda^2} \cdot \sqrt{14}}$

Simplify: $3 = \frac{|5 + 3\lambda|}{\sqrt{5 + \lambda^2}}$

Squaring both sides:
$9 = \frac{(5 + 3\lambda)^2}{5 + \lambda^2}$

Cross-multiply and simplify:
$9(5 + \lambda^2) = 25 + 30\lambda + 9\lambda^2$
$45 + 9\lambda^2 = 25 + 30\lambda + 9\lambda^2$
$20 = 30\lambda$

Solving gives $\lambda = \frac{2}{3}$

Final answer: $\boxed{\text{A}}$