Note that the line is parallel to the plane, hence the image of the line will be a parallel line.

Now,  let the image of $\left(\mathrm{2,3},4\right)$ in $x+y+z=6$ is $\left(x_1,y_1,z_1\right)$

Then, $\frac{x_1-2}{1}=\frac{{\displaystyle y_1-3}}{{\displaystyle 1}}=\frac{{\displaystyle z_1-4}}{{\displaystyle 1}}=-2\left(\frac{2+3+4-6}{1+1+1}\right)$ 

i.e. $\left(x_1,y_1,z_1\right)\equiv \left(\mathrm{0,1},2\right)$
So the required equation is $\overrightarrow{r}=\left(\widehat{j}+2\widehat{k}\right)+\beta \left(2\widehat{i}-3\widehat{j}+\widehat{k}\right)$