1. **State the problem:** Simplify the expression $$-3\sqrt{7r^3} \cdot 6\sqrt{7r^2}$$. 2. **Recall the formula:** The product of square roots is the square root of the product: $$\sqrt{a} \cdot \sqrt{b} = \sqrt{ab}$$. 3. **Apply the formula:** $$-3\sqrt{7r^3} \cdot 6\sqrt{7r^2} = (-3)(6) \sqrt{7r^3 \cdot 7r^2} = -18 \sqrt{49 r^{5}}$$ 4. **Simplify inside the square root:** $$\sqrt{49 r^{5}} = \sqrt{49} \cdot \sqrt{r^{5}} = 7 \cdot \sqrt{r^{5}}$$ 5. **Simplify the radical with exponents:** Recall that $$\sqrt{r^{5}} = r^{\frac{5}{2}} = r^{2} \cdot r^{\frac{1}{2}} = r^{2} \sqrt{r}$$. 6. **Substitute back:** $$-18 \cdot 7 \cdot r^{2} \sqrt{r} = -126 r^{2} \sqrt{r}$$ **Final answer:** $$\boxed{-126 r^{2} \sqrt{r}}$$