1. Hyperbola data

Given $H:\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ and it passes through $(2\sqrt2,-2\sqrt2)$.

So, $\frac{(2\sqrt2)^2}{a^2}-\frac{(-2\sqrt2)^2}{b^2}=1$ $\frac{8}{a^2}-\frac{8}{b^2}=1$ $8\left(\frac{1}{a^2}-\frac{1}{b^2}\right)=1 \quad \text{(Equation 1)}$

For the hyperbola, eccentricity is $e=\frac{c}{a}, \qquad c^2=a^2+b^2$

Length of latus rectum of hyperbola: $L_H=\frac{2b^2}{a}$

2. Parabola construction

The focus of the parabola is the focus of $H$ with positive abscissa, so $F=(c,0)$ The other focus of $H$ is $(-c,0)$.

Let the directrix of parabola be the vertical line $x=d$ Since it passes through $(-c,0)$, we get $d=-c$

Thus parabola has focus $(c,0)$ and directrix $x=-c$. Its axis is the $x$-axis, and vertex is midpoint between focus and directrix: $\left(\frac{c+(-c)}{2},0\right)=(0,0)$ So the parabola is $y^2=4cx$ with parameter $c$.

Length of latus rectum of parabola: $L_P=4c$

3. Using the latus rectum condition

Given: $L_P=e\cdot L_H$ So, $4c=e\cdot \frac{2b^2}{a}$ Using $e=\frac{c}{a}$, $4c=\frac{c}{a}\cdot \frac{2b^2}{a}$ Assuming $c\neq 0$, divide by $c$: $4=\frac{2b^2}{a^2}$ $b^2=2a^2$

4. Find $a,b,c$

Substitute $b^2=2a^2$ into Equation 1: $8\left(\frac{1}{a^2}-\frac{1}{2a^2}\right)=1$ $8\left(\frac{1}{2a^2}\right)=1$ $\frac{4}{a^2}=1$ $a^2=4$ Then $b^2=8$ And $c^2=a^2+b^2=4+8=12$ $c=2\sqrt3$

Hence the parabola is $y^2=4cx=8\sqrt3\,x$

5. Check the options

We test each point in $y^2=8\sqrt3\,x$

Option A: $(2\sqrt3,3\sqrt2)$

$y^2=(3\sqrt2)^2=18$ $8\sqrt3\,x=8\sqrt3(2\sqrt3)=8\cdot 2\cdot 3=48$ $18\neq 48$ So A is not on the parabola.

Option B: $(3\sqrt3,-6\sqrt2)$

$y^2=(-6\sqrt2)^2=72$ $8\sqrt3\,x=8\sqrt3(3\sqrt3)=8\cdot 3\cdot 3=72$ $72=72$ So B lies on the parabola.

Option C: $(\sqrt3,-\sqrt6)$

$y^2=(-\sqrt6)^2=6$ $8\sqrt3\,x=8\sqrt3(\sqrt3)=24$ $6\neq 24$ So C is not on the parabola.

Option D: $(3\sqrt6,6\sqrt2)$

$y^2=(6\sqrt2)^2=72$ $8\sqrt3\,x=8\sqrt3(3\sqrt6)=24\sqrt{18}=72\sqrt2$ $72\neq 72\sqrt2$ So D is not on the parabola.

6. Conclusion

The point lying on the parabola is: $\boxed{(3\sqrt3,-6\sqrt2)}$ which is Option B.