Given that the first term $a$ and common ratio $r$ of a geometric progression be positive integer. So, their 1st three terms are $a, a r, a r^2$

According to the question, $a^2+a^2 r^2+a^2 r^4=33033$

$$ \begin{aligned} \Rightarrow a^2\left(1+r^2+r^4\right) & =3 \times 7 \times 11 \times 11 \times 13 \\ & =3 \times 7 \times 13 \times 11^2 \end{aligned} $$

$$ \begin{aligned} & \therefore \quad a^2=11^2 \\\\ & \Rightarrow \quad a=11 \\\\ & \text { and } 1+r^2+r^4=273 \\\\ & \Rightarrow r^2+r^4=272 \\\\ & \Rightarrow r^4+r^2-272=0 \\\\ & \Rightarrow\left(r^2+17\right)\left(r^2-16\right)=0 \\\\ & \Rightarrow r^2=-17(not ~possible),\\\\ & \Rightarrow r^2-16=0 \\\\ & \text { } \Rightarrow r= \pm 4 \\\\ & \Rightarrow r=4 ~~( \because r > 0) \end{aligned} $$

So, sum of these first three terms is $a+a r+a r^2$ $=11+44+176=231$