Draw the graph according to the question.
According to the graph, the $x$ and $y$ coordinates of the centre is the radius. Then Let $I(r,r)$ be the coordinates of the in-centre.
Since the in-centre of $\Delta OPQ$ lies on the straight line $3x+4y = 3p$, then we have
$\begin{array}{rcl}
3r + 4r & = & 3p \\
7r & = & 3p \\
r & = & \dfrac{3p}{7}
\end{array}$
Consider $\Delta OPQ$.
$\begin{array}{rcl}
QA & = & OQ – OA \\
QA & = & q – r \\
QA & = & q – \dfrac{3p}{7}
\end{array}$
Also,
$\begin{array}{rcl}
PB & = & OP – OB \\
PB & = & p – r \\
PB & = & p – \dfrac{3p}{7} \\
PB & = & \dfrac{4p}{7}
\end{array}$
Since $OP$, $OQ$ and $PQ$ are tangents to the circle, then we have $QC = QA$ and $PC = PB$ (tangent properties). Hence, we have
$\begin{array}{rcl}
PQ & = & PC + QC \\
PQ & = & PB + QA \\
\sqrt{(p-0)^2 + (q-0)^2} & = & \dfrac{4p}{7} + q – \dfrac{3p}{7} \\
p^2 + q^2 & = & (\dfrac{p}{7} +q)^2 \\
p^2 + q^2 & = & \dfrac{p^2}{49} + \dfrac{2pq}{7} + q^2 \\
\dfrac{48p^2}{49} & = & \dfrac{2pq}{7} \\
\dfrac{p^2}{pq} & = & \dfrac{7}{24} \\
\dfrac{p}{q} & = & \dfrac{7}{24} \\
p : q & = & 7 : 24
\end{array}$
