By linear combinations, the two equations can be rearranged as:
x = 5/11 and z – y = 5//11
Thus, the unit vector perpendicular to one plane is (1 , 0, 0)
and the unit vector perpendicular to the other plane is ( 0, -.707 , .707)
I am using .707 in place of 1/sqrt(2) --- easier to type
The cross product of these two unit vectors will be a vector parallel to the line of intersection of the planes.
The cross product is ( 0 , -.707 , -.707 )
From this, a parametric expression for the desired line is
(2, -3 - .707 k , -7 -.707 k) for all real k.
By eliminating k between the y and z components, an alternate form for the line is the two equations
x = 2 z = y -4