Playing with co-ordinate geometry

We have a plane in $xyz$ -Cartesian Space whose equation is given by $x+y+z=1$ .

I now take this plane as ${x}^{'}{y}^{'}$ -Cartesian Plane where ${x}^{'}$ and ${y}^{'}$ -axes are given by $x+y=1=z+1$ and $2x=2y=1-z$ respectively.

The Direction cosines of $\Large +{x}^{'}$ and $+{y}^{'}$ axes are $(\frac{-1}{\sqrt{2}},\frac{1}{\sqrt{2}},0)$ and $(\frac{-1}{\sqrt{6}},\frac{-1}{\sqrt{6}},\frac{2}{\sqrt{6}})$ respectively.

A line is chosen in our given plane whose equation in $xyz$ -Cartesian Space is given by :

$3x-1=\frac{3y-1}{2}=\frac{3z-1}{-3}$

If the equation of the above mentioned line in ${x}^{'}{y}^{'}$ -Cartesian Plane is given by :

$a{x}^{'}+b{y}^{'}=1$

Find $\sqrt{2}a+\sqrt{6}b$