The perimeter of a triangle is

Let $a, b$ and $c$ be the three sides of the triangle.

$\therefore$ Perimeter of the triangle $=(a+b+c)$

Given perimeter of the triangle $=6 p^{2}-4 p+9$

One side $(a)=p^{2}-2 p+1$

Other side $(b)=3 p^{2}-5 p+3$

Perimeter $=(a+b+c)$

$\left(6 p^{2}-4 p+9\right)=\left(p^{2}-2 p+1\right)+\left(3 p^{2}-5 p+3\right)+c$

$6 p^{2}-4 p+9-p^{2}+2 p-1-3 p^{2}+5 p-3=c$

$\left(6 p^{2}-p^{2}-3 p^{2}\right)+(-4 p+2 p+5 p)+(9-1-3)=c$

$2 p^{2}+3 p+5=c$

Thus, the third side is $2 p^{2}+3 p+5$.