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