Write the parabola in standard form

Given parabola: $y^2=12x$ Comparing with $y^2=4ax$, we get $4a=12 \Rightarrow a=3$ So the focus is $S=(3,0)$

Equation of a focal chord with slope $m$

A chord passing through the focus with slope $m$ has equation $y=m(x-3)$ or $y=mx-3m$

Find the points of intersection with the parabola

Substitute $y=mx-3m$ into $y^2=12x$: $\big(m(x-3)\big)^2=12x$ $m^2(x-3)^2=12x$ $m^2(x^2-6x+9)-12x=0$ $m^2x^2-(6m^2+12)x+9m^2=0$

Let the intersection points be $A,B$ with $x$-coordinates $x_1,x_2$.

Then $x_1+x_2=\frac{6m^2+12}{m^2}=6+\frac{12}{m^2}, \qquad x_1x_2=9$

Length of the focal chord

Since both points lie on the line of slope $m$, if their difference in $x$-coordinates is $|x_1-x_2|$, then $AB=|x_1-x_2|\sqrt{1+m^2}$

First compute $|x_1-x_2|$: $(x_1-x_2)^2=(x_1+x_2)^2-4x_1x_2$ $=\left(6+\frac{12}{m^2}\right)^2-36$ $=36+\frac{144}{m^2}+\frac{144}{m^4}-36$ $=\frac{144}{m^2}\left(1+\frac{1}{m^2}\right)$

Hence $|x_1-x_2|=\frac{12}{|m|}\sqrt{1+\frac{1}{m^2}}$

Therefore $l=AB=|x_1-x_2|\sqrt{1+m^2}$

Using the standard focal-chord result for $y^2=4ax$, the length is $l=\frac{4a(1+m^2)}{|m|}=\frac{12(1+m^2)}{|m|}$

Distance of the chord from the origin

The line is $mx-y-3m=0$ Distance from origin $(0,0)$ to this line is $d=\frac{| -3m |}{\sqrt{m^2+1}}=\frac{3|m|}{\sqrt{1+m^2}}$

So $d^2=\frac{9m^2}{1+m^2}$

Compute $ld^2$

$ld^2=\frac{12(1+m^2)}{|m|}\cdot \frac{9m^2}{1+m^2}$ $=108|m|$

Use the condition $m<\sqrt{3}$

For a focal chord of parabola, the two slopes of the focal chord endpoints satisfy $m_1m_2=-1$, but here the chord itself has slope $m$ and the condition $m<\sqrt{3}$ does not by itself fix $m$.

So we re-interpret using the standard parametric form of endpoints of a focal chord: $A=(at_1^2,2at_1),\quad B=(at_2^2,2at_2),\quad t_1t_2=-1$ with $a=3$.

Let $t_1=t,\ t_2=-1/t$.

Then slope of chord $AB$ is $m=\frac{2a(-1/t-t)}{a(1/t^2-t^2)}=\frac{2t}{t^2-1}$

Length of focal chord: $l=a\left(t+\frac{1}{t}\right)^2=3\left(t+\frac{1}{t}\right)^2$

Equation of chord joining $(3t^2,6t)$ and $(3/t^2,-6/t)$ simplifies to $tx+y-t(1-t^2)=0$ but using the direct line through focus with slope $m$ is simpler, giving $d=\frac{3|m|}{\sqrt{1+m^2}}$

Since $m=\frac{2t}{t^2-1}$ one gets $1+m^2=\frac{(t^2+1)^2}{(t^2-1)^2}$ and hence $d^2=\frac{36t^2}{(t^2+1)^2}$

Therefore $ld^2=3\left(t+\frac{1}{t}\right)^2\cdot \frac{36t^2}{(t^2+1)^2}$ $=108\cdot \frac{(t^2+1)^2}{t^2}\cdot \frac{t^2}{(t^2+1)^2}$ $=108$

Final answer

$\boxed{108}$