 Solution

Denote the columns by $c_1,\dots ,c_4$. The first two are linearly independent, because the top two coordinates have determinant $3\cdot 2-(-1)\cdot 1=7$. Direct calculation givesThus a column-space basis is $\{(3,1,9{)}^T,(-1,2,4{)}^T\}$, andFor the last description, the third row equals twice the first row plus three times the second, so every column lies in that plane. Since the two independent columns already span a two-dimensional vector subspace, they fill the plane.

The target vector satisfies $5=2(-2)+3(3)$, so it is in the column space. An especially simple linear combination isIn terms of the displayed basis, the same vector is $-c_1/7+11c_2/7$. The column space itself is unique; the freedom is in choosing its basis.