Consider a simpler question: How to find the basis for the subspace of R2 defined by the equation 2x - y = 0. In this case you can solve for a variable and parametrize. We have y = 2x, and we let x = t. So <x, y> = <t, 2t> = t <1, 2>, and a basis for the (one dimensional) subspace is <1,2>.
Your subspace will be three dimensional. Solve for x1 in terms of the other variables and parametrize them. We have x1 = (1/-6) * (5x2 - 8x3 + 7x4). Replacing the other variables with parameters s, t, and u respectively, we have a system, x1 = (-5/6)s + (8/6)t + (-7/6)u, x2 = s, x3 = t, x4 = u.
Altogether then,
<x1, x2, x3, x4> = s<-5/6, 1, 0, 0> + t<8/6, 0, 1, 0> + u<-7/6, 0, 0, 1>
The three vectors on the right hand side form a basis for your subspace. (You can scale them by 6 to remove fractions, if you prefer.)
