It sounds like you might be in a linear algebra course or maybe you're taking an advanced algebra or precalculus course. There are several different expected ways to answer this question depending on which course you're taking and the material you've learned
so far. A cumbersome-but-most-likely-to-be-familiar solution method is to avoid vectors entirely by converting the vector equation into a system of three equations and three unknowns:
x1 + 4x2 + 7x3 = 0
2x1 + 5x2 + 8x3 = 0
3x1 + 6x2 + 9x3 = 0
Solve the top equation for x1 in terms of x2 and x3. Next, use that equation by substituting for x1 in the bottom two equations. The end result will be that you'll have two equations with two unknowns, but they're
dependent equations. They'll describe the same line. You can see that by multiplying both sides of one equation by a constant to obtain the other one.
You should be able to determine that both equations predict that x2 = –2x3 . Substituting this equality back into your equation for x1 in terms of x2 and x3 will yield x1 = x3. And that's it. You have
an infinite number of possible solutions.
Pick any three numbers where the first and third are the same and the second is –2 times the others, and you have values for your x's that work. The answer you listed is a solution but it's uglier than the combination where x1=x3=1
and x2=–2. With these choices, w1 – 2w2 +w3 also equals the zero vector, it's much prettier, and it's the sort of answer you'll be expected to give soon if you're taking a linear algebra course once you begin working
Related to the matrix equation, if you're learning about determinants, invertibility, and linear dependence then these may be useful:
If your linear algebra course is moving on to eigensystems soon then this is an example of finding an eigenvector corresponding to an eigenvalue of zero.