The symmetric form of the line of intersection of two planes $x-y+2z=5,3x+y+z=7$ is

The given planes are $x-y+2z=5,3x+y+z=7$

Consider the normal vectors and then find the cross product of those two vectors to get the vector along the line of intersection of the given two planes

$n_1\times n_2=\left|\begin{array}{ccc}i& j& k\\ 1& -1& 2\\ 3& 1& 1\end{array}\right|=i(-3)-j(-5)+k\left(4\right)$

Hence the direction ratios of the required line are proportional to $⟨3,-5,-4⟩$

To get the point of intersection, plug in $z=0$ in the two plane equations and then solve for $x,y$ 

$x-y=5,3x+y=7$

Adding these two equations and then simplify  $x=3,y=-2$

Therefore, the equation of the line is $\boxed{\frac{x-3}{3}=\frac{y+2}{-5}=\frac{z}{-4}}$