If (2, -1, 2, 1) is a linear combination of the three given vectors, then there should exist [tex]c_1,c_2,c_3[/tex] such that
[tex](2,-1,2,1)=c_1(1,-2,0,3)+c_2(2,3,0,-1)+c_3(3,9,-4,-2)[/tex]
or equivalently, there should exist a solution to the system
[tex]\begin{cases}c_1+2c_2+3c_3=2\\-2c_1+3c_2+9c_3=-1\\-4c_3=2\\3c_1-c_2-2c_3=1\end{cases}[/tex]
Right away we get [tex]c_3=-\dfrac12[/tex], so the system reduces to
[tex]\begin{cases}c_1+2c_2=\dfrac72\\\\-2c_1+3c_2=\dfrac72\\\\3c_1-c_2=0\end{cases}[/tex]
Notice that the first equation is the sum of the latter two. The third equation gives us
[tex]3c_1-c_2=0\implies 3c_1=c_2[/tex]
so that in the second equation,
[tex]-2c_1+3c_2=\dfrac72\implies7c_1=\dfrac72\implies c_1=\dfrac12[/tex]
which in turn gives
[tex]3c_1=c_2\implies c_2=\dfrac32[/tex]
and hence the (2, -1, 2, 1) is a linear combination of the given vectors, with
[tex]\boxed{(2,-1,2,1)=\dfrac12(1,-2,0,3)+\dfrac32(2,3,0,-1)-\dfrac12(3,9,-4,-2)}[/tex]
