Is the vector $(3,-1,0,-1)$ in the subspace of $\Bbb R^4$ spanned by the vectors $(2,-1,3,2)$, $(-1,1,1,-3)$, $(1,1,9,-5)$? Is the vector $(3,-1,0,-1)$ in the subspace of $\Bbb R^4$ spanned by the vectors $(2,-1,3,2)$, $(-1,1,1,-3)$, $(1,1,9,-5)$?
To find wether $(3,-1,0,-1)$ is in the span of the other vectors, solve the system:
$$(3,-1,0,-1)=\lambda_1(2,-1,3,2)+\lambda _2(-1,1,1,-3)+\lambda _3(1,1,9,-5)$$

If you get a solution, then the vector is the span. If you don't get a solution, then it isn't.

It's worth noting that the span of $(2,-1,3,2), (-1,1,1,-3), (1,1,9,-5)$ is exactly the set $\left\{\lambda_1(2,-1,3,2)+\lambda _2(-1,1,1,-3)+\lambda _3(1,1,9,-5):\lambda _1, \lambda _2, \lambda _3\in \Bbb R\right\}$



Alternatively you can consider the matrix $\begin{bmatrix}2&amp; -1 &amp;3 &amp; 2\\ -1 &amp;1 &amp;1 &amp;-3\\ 1 &amp; 1 &amp; 9 &amp; -5 \\ 3 &amp;-1 &amp;0 &amp; -1\end{bmatrix}$. Compute its determinant. If it's not $0$, then the four vectors are linearly independent. If it is $0$ they are linearly dependent. What does that tell you?
if (3,-1,0,-1) be in span {(2,-1,3,2),(-1,1,1,-3),(1,1,9,-5)} then we must have
 (3,-1,0,-1)=$c_{1}$ (2,-1,3,2)+ $c_2$ (-1,1,1,-3)+ $c_3$ (1,1,9,-5)
so this system must have solution:

$2c_1 -c_2+c_3=3$

$-c_1 +c_2+c_3=-1$

$3c_1 +c_2+9c_3=0$

$2c_1 -3c_2-5c_3=-1$ 
since 3 first relation has unic answer and 3 end realations too and they cant have equal answer (then this system has not solution and (3-10-1)