[applied maths] vector

d = xa + yb + zc
7i + 23j + 29k=x(3i + 4j +5k)+y(2i + 2j + 3k)+z(6i -7j -8k)
7i + 23j + 29k=(3x+2y+6z)i +(4x+2y-7z)j+(5x+3y-8z)k

So
3x+2y+6z=7
4x+2y-7z=23
5x+3y-8z=29

Using Gauss elimination, we have x=3,y=2,z=-1