Let v=xu1+yu2+zu3. Then
(1,−2,5)=x(1,1,1)+y(1,2,3)+z(2,−1,1)=(x+y+2z,x+2y−z,x+3y+z),
and hence we get the following system
⎩⎨⎧x+y+2z=1x+2y−z=−2x+3y+z=5
which is equivalent (after substracting from the second row the first row, from the third row the second row) to
⎩⎨⎧x+y+2z=1y−3z=−3y+2z=7
and (after substracting from the third row the second row) to
⎩⎨⎧x+y+2z=1y−3z=−35z=10
It follows that z=2, y=3z−3=6−3=3, and x=1−y−2z=1−3−4=−6.
We conclude that v=−6u1+3u2+2u3.
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