Given,

The equation of the plane containing the line $x+2y+3z-4=0=2x+y-z+5$ and perpendicular to the plane$\overrightarrow{r}=\left(\hat{i}-\hat{j}\right)+\lambda \left(\hat{i}+\hat{j}+\hat{k}\right)+\mu (\hat{i}-2\hat{j}+3\hat{k})$ is $ax+by+cz=4$, 

Now let equation of required plane be,

$P: \left(x+2y+3z-4\right)+\alpha \left(2x+y-z+5\right)=0$

$\Rightarrow (2\alpha +1)+\left(\alpha +2\right)y+\left(3-\alpha \right)z=4-5\alpha$

So, the normal vector is given by,

$\overrightarrow{n_1}=(2\alpha +1)\hat{i}+(\alpha +2)\hat{j}+(3-\alpha )\hat{k}$

And normal of the plane $\overrightarrow{r}=\left(\hat{i}-\hat{j}\right)+\lambda \left(\hat{i}+\hat{j}+\hat{k}\right)+\mu (\hat{i}-2\hat{j}+3\hat{k})$ will be, 

$\overrightarrow{n_2}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 1& 1& 1\\ 1& -2& 3\end{array}\right|=5\hat{j}-2\hat{j}-3\hat{k}$

Now given they are perpendicular, so by perpendicular condition we get,

$\overrightarrow{n_1}·\overrightarrow{n_2}=0$

$\Rightarrow 5\left(2\alpha +1\right)-2\left(\alpha +2\right)-3\left(3-\alpha \right)=0$

$\Rightarrow 11\alpha +\left(-8\right)=0$

$\Rightarrow \alpha =\frac{8}{11}$

Hence, equation of required plane will be,

$P:\frac{27}{11}x+\frac{30}{11}y+\frac{25}{11}z-\frac{4}{11}=0$

$\Rightarrow 27x+30y+25z=4$

Now comparing with $ax+by+cz=4$ we get,

$a=27, b=30, c=25$

Hence, $a-b+c=22$