Question 1026390
The normal vector defining the equation of the plane is {3, -7, 5}.
Let (x,y,z) be a point on the desired line.  Then combined with the given point (2,-1,3) the vector {x-2, y+1, z-3} must be parallel to the vector {3, -7, 5}.
Hence {x-2, y+1, z-3}  = k{3, -7, 5}

==> the symmetric equation would be {{{(x-2)/3 = (y+1)/-1 = (z-3)/3}}}