Answer
$f_{1},f_{2},f_{3}$ are linearly independent.
Work Step by Step
$af_{1}(t)+bf_{2}(t)+cf_{3}(t)=\;a(2t-3)\;+b(t^2+1)\;+c(2t^2-t)=\;0\\\\$
$(-3a+b)+\;(2a-c)t+(b+2c)t^2=0\\\\$
By the polynomial equality theorem:
$-3a+b=0 \;\;\;\;\;\;\;\;\; \rightarrow \;\;\;\;\boxed{b=3a}\\\\$
$2a-c=0 \;\;\;\;\;\;\;\;\; \rightarrow \;\;\;\;\boxed{c=2a}\\\\$
$b+2c=0 \;\;\;\;\;\;\;\; \rightarrow \;\;\;\;(3a)+2(2a)=0\;\;\;\;\;\rightarrow 7a=0\;\;\; \rightarrow
\color{Red} {\boxed {a=0}}\\\\$
$a=0 \;\;\;\;\;\;\;\; b=0 \;\;\;\;\;\;\;\; c=0$
so, $f_{1},f_{2},f_{3}$ are linearly independent.
