The given equations are,

$L:\frac{x-0}{1}=\frac{y-1}{2}=\frac{z-3}{\lambda } \& P:x+2y+3z=4$

So, Vector parallel to line$:<1, 2, \lambda >=\overrightarrow{b}$

Normal vector to plane $P:<1, 2, 3>=\overrightarrow{n}$

Angle between plane & line is $\theta$. Then angle between the line and normal to the plane will be $90-\theta$.
Then, $\Rightarrow \cos \left(90-\theta \right)=\frac{\left(\hat{i}+2\hat{j}+\lambda \hat{k}\right)·\left(\hat{i}+2\hat{j}+3\hat{k}\right)}{\sqrt{1^2+2^2+{\lambda }^2}·\sqrt{1^2+2^2+3^2}}$
$\Rightarrow \frac{3}{\sqrt{14}}=\frac{1+4+3\lambda }{\sqrt{{\lambda }^2+5}\sqrt{14}}\Rightarrow \lambda =\frac{2}{3}$

$L_1=\frac{x-0}{3}=\frac{y-1}{6}=\frac{z-3}{2}=\mu$

$\left(x,y,z\right)\equiv \left(3\mu ,6\mu +1,2\mu +3\right)$

But this point lies on the plane :

$\left(3\mu +12\mu +2+6\mu +9\right)=4$

$\Rightarrow \mu =\frac{-1}{3}$
Hence, $\alpha =3\mu =-1, \beta =6\mu +1=-1, \gamma =2\mu +3=\frac{7}{3}$

Now, $\alpha +2\beta +6\gamma =11$

Hence this is the required answer.