Answer
See below
Work Step by Step
Consider set $\{(1,-1,2,3),(2,-1,1,-1),(-1,1,1,1)$
Let $a,b,c$ be scalars such that
$a(1,-1,2,3)+b(2,-1,1,-1)+c(-1,1,1,1)=(0,0,0,0)\\
(a+2b-c,-a-b+c,2a+b+c,3a-b+c)=(0,0,0,0)$
We have the system: $a+2b-c=0\\
-a-b+c=0\\
2a+b+c=0\\
3a-b+c=0$
then $a+2b-c+(-a-b+c)=0\rightarrow b=0$
Substitute $a-c=0\\
-a+c=0\\
\rightarrow a=c$
and $3a-b+c=3a-0+a=2a=0 \rightarrow a=0\\
\rightarrow c=0$
Hence, the linear combination of $\{(1,-1,2,3),(2,-1,1,-1),(-1,1,1,1)$ is the trivial one and the set $\{(1,-1,2,3),(2,-1,1,-1),(-1,1,1,1)$ is linearly independent in $R^4$
