adding #1 and #2 I got 2x + 5z = 6
adding #1 and #3 I got 4x + 10z = 13
You should realize that we cannot solve this any further, since both x and z would drop out
so lets find the line of intersection of plane #1 and #2
from 2x + 5z = 6
let z=2, then x = -2 and from #2, y = 0
let z=4, then x = -7 and from #2 y = 1
so we have two points on the line of intersection of the first two planes, namely
(-2,0,2) and (-7,1,4) from which we can get a direction vector (5,-1,-2)
and parametric equations of that line could be
x = -2+5t
y = -t
z = 2-2t
subbing that into the third equation 3x + y + 7z = 0 we get
-6 + 15t - t + 14 - 14t = 9
8=9 which is a contradiction.
Had this been a TRUE statement then all 3 planes would have intersected in the same line, like the spine of an open book
but the statement was FALSE, so the 3 planes must intersect in 3 different but parallel lines,
to find the other two line equations follow my procedure above for the other two possible pairs of planes.
As a check they should all have the same direction numbers.
BTW, this question is the hardest of the possible types of intersection of 3 planes.