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 \(\sqrt{2}a+\sqrt{6}b\)

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