Show that every vector $b\in \mathbb{R}^3$ lies in $span \left((1,2,3),(-1,-1,0),(2,1,-1)\right)$

Question

Show that every vector $b\in \mathbb{R}^3$ lies in

$$span \left((1,2,3),(-1,-1,0),(2,1,-1)\right)$$

My attempt:

suppose $b=(b_1,b_2,b_3)=x(1,2,3)+y(-1,-1,0)+z(2,1,-1)$

then $\;x-y+2z=b_1\\ 2x+-y+z=b_2\\ 3x-z=b_3 $

if the system have soltion then every vector $b\in \mathbb{R}^3$ lies in

$span \left((1,2,3),(-1,-1,0),(2,1,-1)\right)$

Answer 1

Reduce the matrix in RREF:$$\left( \begin{array}{ccc} 1 & -1 & 2 \\ 2 & -1 & 1 \\ 3 & 0 & -1 \\ \end{array} \right)\sim \left( \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\\ \end{array} \right)$$ the system will be consistent, therefore $b=\text{span} \left\{(1,2,3),(-1,-1,0),(2,1,-1)\right\}.$

Answer 2

All you need to do is run Gram-Schmidt on the set of vectors you have and show that it gives you a set of 3 orthoogonal vectors. Row-reduction also works.