The span of 3 vectors can have dimension at most 3, so 9 is certainly not correct.
Check whether the 3 vectors are linearly independent. If they are not, then there is some choice of scalars [tex]c_1,c_2,c_3[/tex] (not all zero) such that
[tex]c_1 (2,-1,1) + c_2 (3,1,1) + c_3 (1,2,0) = (0,0,0)[/tex]
which leads to the system of linear equations,
[tex]\begin{cases} 2c_1 + 3c_2 + c_3 = 0 \\ -c_1 + c_2 + 2c_3 = 0 \\ c_1 + c_2 = 0 \end{cases}[/tex]
From the third equation, we have [tex]c_1=-c_2[/tex], and substituting this into the second equation gives
[tex]-c_1 + c_2 + 2c_3 = 2c_2 + 2c_3 = 0 \implies c_2 + c_3 = 0 \implies c_2 = -c_3[/tex]
and in turn, [tex]c_1=c_3[/tex]. Substituting these into the first equation gives
[tex]2c_1 + 3c_2 + c_3 = 2c_3 - 3c_3 + c_3 = 0 \implies 0=0[/tex]
which tells us that any value of [tex]c_3[/tex] will work. If [tex]c_3 = t[/tex], then [tex]c_1=t[/tex] and [tex]c_2 = -t[/tex]. Therefore the 3 vectors are not linearly independent, so their span cannot have dimension 3.
Repeating the calculations above while taking only 2 of the given vectors at a time, we see that they are pairwise linearly independent, so the span of each pair has dimension 2. This means the span of all 3 vectors taken at once must be 2.
