The equation of line parallel to the plane $x+y+z=2$ and intersecting the line $x+y-z=0=x+2y-3z+5$ at the point  $\left(-4,1,3\right)$ is

Solution:

The equation of plane is $x+y-z+\lambda (x+2y-3z+5)=0$

It is passing through the point $\left(-4,1,3\right)$

It implies that$-4+1-3+\lambda (-4+2-9+5)=0\Rightarrow \lambda =-1$

Hence, the equation of plane is  $y-2z+5=0$

The required line is parallel to both planes $y-2z+5=0$ and $x+y+z=2$

Normal vectors to above planes are$n_1=i+j+k,n_2=j-2k$  

The vector along the line is perpendicular to both the vectors, so it is along  $n_1\times n_2$

Hence,$n_1\times n_2=\left|\begin{array}{ccc}i& j& k\\ 1& 1& 1\\ 0& 1& -2\end{array}\right|=-3i+2j+k$

Hence, the direction ratios of the required line are $⟨-3,2,1⟩$

 Therefore, the equation of the required line is  $\boxed{\frac{x+4}{-3}=\frac{y-1}{2}=\frac{z-3}{1}}$