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

$L: \frac{x-0}{1}=\frac{y-1}{2}=\frac{z-3}{\lambda} $ <br/><br/>$ P: x+2 y+3 z=4$ <br/><br/>Vector parallel to line : $\langle 1,2, \lambda\rangle=\bar{b}$ <br/><br/>Normal vector to plane $P:<1,2,3\rangle=\bar{n}$ <br/><br/>Angle between plane and line is $\theta$ <br/><br/>Then, $\sin \theta=\frac{<1,2, \lambda>\cdot<1,2,3>}{\sqrt{1^2+2^2+\lambda^2} \cdot \sqrt{1^2+2^2+3^2}}$ <br/><br/>$$ \Rightarrow \frac{3}{\sqrt{14}}=\frac{1+4+3 \lambda}{\sqrt{\lambda^2+5} \sqrt{14}} \Rightarrow \lambda=\frac{2}{3} $$ <br/><br/>$L_1 \equiv \frac{x-0}{3}=\frac{y-1}{6}=\frac{z-3}{2}=\mu$ <br/><br/>Any point on line : $(3 \mu, 6 \mu+1,2 \mu+3)$ <br/><br/>It lies on P <br/><br/>$$ \begin{aligned} & \therefore 3 \mu+12 \mu+2+6 \mu+9=4 \\\\ & \Rightarrow \mu=\frac{-1}{3} \end{aligned} $$ <br/><br/>Hence, $\alpha=3 \mu=-1, \beta=6 \mu+1=-1, \gamma=2 \mu+3=\frac{7}{3}$ <br/><br/>Now, $\alpha+2 \beta+6 \gamma=11$