Question 1114443
The normals for the plane are (4,3,7) and (2,1,3).
Find the cross product of these two vectors, it's parallel to the line of intersection,
{{{axb=(matrix(3,3,i,j,k,
4,3,7,2,1,3))}}}
{{{axb=(3*3-1*7)i+(7*2-3*4)j+(4*1-2*3)k}}}
{{{axb=(9-7)i+(14-12)j+(4-6)k}}}
{{{axb=(9-7)i+(14-12)j+(4-6)k}}}
{{{axb=(2)i+(2)j+(-2)k}}}
(2,2,-2)
Solve for any point on the line using the equations. 
You can set {{{z=0}}} and solve for x and y of a point,
{{{4x+3y=12}}}
{{{2x+y=4}}}
leads to,
{{{4x+3y-4x-2y=12-8}}}
{{{y=4}}}
So then,
{{{2x+4=4}}}
{{{x=0}}}
So (0,4,0) is one point on the line and with the results from above,
{{{x=0+2t}}}
{{{y=4+2t}}}
{{{z=0-2t}}}
.
.
{{{x=2t}}}
{{{y=4+2t}}}
{{{z=-2t}}}