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

The general form of the equaiton of line is $x-y+2z-5=3x+y+z-7$

Step 1: Find one of the points of intersection of planes

substittute $z=0$ and then add both equations

$4x-12=0\Rightarrow x=3$

Substitute $x=3$ in $x-y=5\Rightarrow y=-2$

Hence, one of the points of intersection of two planes is $\left(3,-2,0\right)$

Step 2: Finding direciton ratios of the line of intersection of two planes

Required line is along the vector which is perpendicular to both normals to the planes

normal vectors to the planes are $\stackrel{\rightharpoonup }{n_1}=i-j+2k,\stackrel{\rightharpoonup }{n_2}=3i+j+k$

Normal vector to these two vectors is $\stackrel{\rightharpoonup }{n_1}\times \stackrel{\rightharpoonup }{n_2}$

Hence, 

       $\begin{array}{rcl}\stackrel{\rightharpoonup }{n_1}\times \stackrel{\rightharpoonup }{n_2}& =& \left|\begin{array}{ccc}i& j& k\\ 1& -1& 2\\ 3& 1& 1\end{array}\right|\\ & =& i\left(-1-2\right)-j\left(1-6\right)+k\left(1+3\right)\\ & =& -3i+5j+4k\end{array}$

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