Let a curve $y=f(x), x \in(0, \infty)$ pass through the points $P\left(1, \frac{3}{2}\right)$ and $Q\left(a, \frac{1}{2}\right)$. If the tangent at any point $R(b, f(b))$ to the given curve cuts the $\mathrm{y}$-axis at the point $S(0, c)$ such that $b c=3$, then $(P Q)^{2}$ is equal to __________.

Equation of tangent at $R(b, f(b))$ <br/><br/>$y-f(b)=f^{\prime}(b)(x-b)$ <br/><br/>which passes through $S(0, c)$ <br/><br/>$$ \begin{aligned} & \therefore c-f(b)=f^{\prime}(b)(0-b) \\\\ & b f^{\prime}(b)-f(b)=-c \\\\ & \Rightarrow b f^{\prime}(b)-f(b)=\frac{-3}{b} (\because b c=3) \\\\ & \Rightarrow \frac{b f^{\prime}(b)-f(b)}{b^2}=\frac{-3}{b^3} \\\\ & \Rightarrow d\left(\frac{f(b)}{b}\right)=\frac{-3}{b^3} \\\\ & \Rightarrow \frac{f(b)}{b}=\frac{3}{2 b^2}+c \end{aligned} $$ <br/><br/>which passes through $P\left(1, \frac{3}{2}\right)$ <br/><br/>$$ \begin{aligned} & \Rightarrow \frac{3 / 2}{1}=\frac{3}{2}+c \\\\ & \Rightarrow c=0 \\\\ & \therefore f(b)=\frac{3}{2 b^2} \times b \\\\ & \Rightarrow f(b)=\frac{3}{2 b} \end{aligned} $$ <br/><br/>$\because$ It passes through $Q\left(a, \frac{1}{2}\right)$ <br/><br/>$$ \begin{aligned} & \therefore \frac{1}{2}=\frac{3}{2 a} \\\\ & \Rightarrow a=3 \\\\ & \therefore P \equiv\left(1, \frac{3}{2}\right) \text { and } Q \equiv\left(3, \frac{1}{2}\right)\\\\ & \therefore (P Q)^2=(3-1)^2+\left(\frac{1}{2}-\frac{3}{2}\right)^2=4+1=5 \end{aligned} $$