is, of course, 6 dimensional and, generally, each equation reduces the dimension by 1 so your subspace is 6- 2= 4 dimensional and you need a basis containing 4 vectors.

Here's a good way of finding that basis. You are given that x1= 2x2 and x3= 7x6. That means that you are not "free" to choose x1 and x3 to be whatever you want- they are fixed by the values of x2 and x6. But all the othersare"free". Take those to be, one at a time, equal to 1 while the other "free variables" are 0.

For example, if x2= 1, x4= x5= x6= 0 (I have skipped x1 and x3 because they are not "free") then x1= 2x2= 2 and x3= 7x6= 0 so we have <2, 1, 0, 0, 0, 0> as one basis vector. If x4= 1, x2= x5= x6, then x1= 2x2= 0 and x3= 7x6= 0 so we have <0, 0, 0, 1, 0, 0> as another basis vector. If x5= 1, x2= x4= x6= 0, then x1= 2x2= 0 and x3= 7x6= 0 so we have <0, 0, 0, 0, 1, 0> as a third basis vector. Finally, if x6= 1, x2= x4= x5= 0, then x1= 2x2= 0 and x3= 7x6= 7 so we have <0, 0, 7, 0, 0, 1> as the fourth basis vector.