first of all, U+W is a direct sum if and only if: dim(U∩W) = 0 (which is the same thing as saying U∩V = {0}).

presumably you mean "direct sum decomposition of R^{3}", but that is not what you wrote.

you cannot just arbitrarily "throw one vector out", you have to verify that the remaining 3 are linearly independent!

for example, {(1,0,0),(0,1,0),(0,0,1),(2,0,0)} is a linearly dependent set, but throwing out (0,1,0) won't fix this.

since you have deduced that dim(U∩V) = 1, we just need to find one non-zero vector in it, to use as a basis.

so suppose a(0,2,1) + b(1,2,0) = c(5,4,-5) + d(0,1,0), that is:

(b,2a+2b,a) = (5c,4c+d,-5c).

then b = 5c = -a, so:

2a+2b = 0 = 4c+d, so d = -4c = (4/5)(-5c) = (4/5)a. this means if we pick a, everything else falls into place. try a = -1.