Any general point on the line $\frac{x-1}{2}=\frac{y+3}{5}=\frac{z+1}{2}$ is $\left(2\lambda +\mathrm{1,5}\lambda -\mathrm{3,2}\lambda -1\right)$
It must satisfy $x-y+z=4\Rightarrow 2\lambda +1-5\lambda +3+2\lambda -1=4$
$\Rightarrow -\lambda +3=4\Rightarrow \lambda =-1$
So, the point of intersection of the plane and the line is $\left(-1,-8,-3\right)$
This point also lies on $ax-2y+bz=3\Rightarrow -a+16-3b=3$
$\Rightarrow a+3b=13\dots .\left(\mathrm{i}\right)$
This point also satisfy $\frac{x-a}{2}=\frac{y-b}{3}=\frac{z-c}{4}$
$\Rightarrow \frac{1+a}{2}=\frac{8+b}{3}=\frac{3+c}{4}\dots \left(\mathrm{i}\mathrm{i}\right)$
By taking $\frac{1+a}{2}=\frac{8+b}{3}\Rightarrow 3+3a=16+2b$
$\Rightarrow 3a-2b=13\dots \left(\mathrm{i}\mathrm{i}\mathrm{i}\right)$
From $\left(\mathrm{i}\right)$ and $\left(\mathrm{i}\mathrm{i}\mathrm{i}\right)$
$a=\frac{65}{11},b=\frac{26}{11}$
From $\left(\mathrm{i}\mathrm{i}\right)$
$\frac{1+\frac{65}{11}}{2}=\frac{8+\frac{26}{11}}{3}=\frac{3+c}{4}$
$\frac{38}{11}=\frac{38}{11}=\frac{3+c}{4}\Rightarrow c=\frac{119}{11}$
$\Rightarrow 11\left(a+b+c\right)=210$