1. Write the parabola in standard form

   Given parabola: $y^2=12x$

   Compare with the standard form: $y^2=4ax$ so, $4a=12 \implies a=3$

   Hence the focus is: $S=(a,0)=(3,0)$

2. Use the fact that $PQ$ is a focal chord

   A focal chord is any chord passing through the focus.

   Let the endpoints of the focal chord be points on the parabola corresponding to parameters $t_1$ and $t_2$.

   For the parabola $y^2=4ax$, a point with parameter $t$ is: $\big(at^2,\,2at\big)$

   So here, points are: $P=(3t_1^2,6t_1), \qquad Q=(3t_2^2,6t_2)$

   For a focal chord of a parabola, the parameters satisfy: $t_1t_2=-1$

3. Use the length formula for focal chord

   For parabola $y^2=4ax$, length of chord joining parameters $t_1,t_2$ is: $PQ=a|t_1-t_2|\sqrt{(t_1+t_2)^2+4}$

   Since $t_1t_2=-1$, let $t_2=-\frac1{t_1}= -\frac1t$.

   Then endpoints are: $P=(3t^2,6t), \qquad Q=\left(\frac{3}{t^2},-\frac{6}{t}\right)$

   A standard result for focal chord length is: $PQ=a\left(t+\frac1t\right)^2$

   Let us verify quickly: [ \Delta x=3\left(t^2-\frac1{t^2}\right), \qquad \Delta y=6\left(t+\frac1t\right) ] so [ PQ^2=9\left(t^2-\frac1{t^2}\right)^2+36\left(t+\frac1t\right)^2 ] Using $t^2-\frac1{t^2}=\left(t+\frac1t\right)\left(t-\frac1t\right)$ and simplifying gives $PQ=3\left(t+\frac1t\right)^2$

   Given $PQ=15$, therefore $3\left(t+\frac1t\right)^2=15$ $\left(t+\frac1t\right)^2=5$

4. Find the equation of the focal chord

   The chord joining points with parameters $t$ and $-\frac1t$ on $y^2=4ax$ is: $y=mx+c$ where slope is

   =\frac{-6\left(t+\frac1t\right)}{3\left(\frac1{t^2}-t^2\right)} =\frac{2t}{t^2-1}$$ But an easier standard form of focal chord is: $$x-ty+at^2=0$$ for one endpoint parameter $t$ and the other $-1/t$. For $a=3$, the focal chord is: $$x-ty+3=0$$ Rearranging: $$x-ty+3=0$$

5. Distance of this line from the origin

   Distance from origin $(0,0)$ to line $x-ty+3=0$ is $p=\frac{|3|}{\sqrt{1+t^2}}=\frac{3}{\sqrt{1+t^2}}$

   So, $p^2=\frac{9}{1+t^2}$

6. Use focal chord length condition to get $p$

   From $\left(t+\frac1t\right)^2=5$ we get $t^2+2+\frac1{t^2}=5$ $t^2+\frac1{t^2}=3$

   Multiply by $t^2$: $t^4-3t^2+1=0$

   Let $u=t^2$. Then $u^2-3u+1=0$

   So, $u+\frac1u=3$ where $u=t^2$.

   Now $p^2=\frac{9}{1+u}$

   We need a value independent of which root is chosen. Since the two roots are reciprocal, the two possible distances are complementary. Let us compute directly using the perpendicular distance formula from the chord in terms of its length.

7. A cleaner method using midpoint form

   For parabola $y^2=4ax$, focal chord endpoints are $P(at^2,2at),\quad Q\left(\frac{a}{t^2},-\frac{2a}{t}\right)$ and its equation is $tx-y+at=0$ or equivalently for the focal chord through parameters $t$ and $-1/t$, $x-ty+a=0$

   Distance from focus $(a,0)$ to the endpoints along this line gives the chord length relation $PQ=\frac{4a(1+t^2)}{|t|}$ This is not matching the known standard formula, so let us proceed by direct coordinate geometry to avoid ambiguity.

8. Direct coordinate method

   Let the focal chord through focus $(3,0)$ have slope $m$. Equation of line through focus: $y=m(x-3)$

   Substitute into parabola $y^2=12x$: $m^2(x-3)^2=12x$

   Since the chord passes through the focus, the two intersection points are $P,Q$, and their distance along the line is given as 15.

   Write line in parametric form from focus: $(x,y)=(3,0)+\lambda(1,m)$ so $x=3+\lambda,\quad y=m\lambda$

   Substituting into parabola: $(m\lambda)^2=12(3+\lambda)$ $m^2\lambda^2-12\lambda-36=0$

   Let roots be $\lambda_1,\lambda_2$. Then points are at parameters along direction vector $(1,m)$.

   Distance between the points is: $PQ=|\lambda_1-\lambda_2|\sqrt{1+m^2}=15$

   For the quadratic $m^2\lambda^2-12\lambda-36=0$ we have

   =\frac{12\sqrt{1+m^2}}{m^2}$$ Hence $$PQ=\frac{12(1+m^2)}{m^2}=15$$ So, $$12+12m^2=15m^2$$ $$3m^2=12$$ $$m^2=4$$

9. Find distance of the line from origin

   The line is: $y=m(x-3)$ $mx-y-3m=0$

   Distance from origin to this line is: $p=\frac{| -3m |}{\sqrt{m^2+1}}=\frac{3|m|}{\sqrt{m^2+1}}$

   Since $m^2=4$, $p^2=\frac{9\cdot 4}{5}=\frac{36}{5}$

   Therefore, $10p^2=10\cdot \frac{36}{5}=72$

10. Final answer

$\boxed{72}$