Let $\Delta ABC$ be an acute-angled triangle and let $H$ be its orthocentre. Let $G_1, G_2$ and $G_3$ be the centroids of the triangles $\Delta HBC , \Delta HCA$ and $\Delta HAB$ respectively. If the area of $\Delta G_1G_2G_3$ is $7$ units, what is the area of $\Delta ABC $? 
1. Given that $\Delta ABC$ is an acute-angled triangle and $H$ is its orthocenter. The centroids of the triangles $\Delta HBC$, $\Delta HCA$, and $\Delta HAB$ are $G_1$, $G_2$, and $G_3$ respectively.
2. We need to find the relationship between the area of $\Delta G_1G_2G_3$ and the area of $\Delta ABC$.

3. For any point $H$ inside the triangle $\Delta ABC$, the area of the triangle formed by the centroids of the triangles $\Delta HBC$, $\Delta HCA$, and $\Delta HAB$ is given by:
   \[
   \frac{[G_1G_2G_3]}{[MNP]} = \left(\frac{2}{3}\right)^2
   \]
   where $M$, $N$, and $P$ are the midpoints of the sides of $\Delta ABC$.

4. The area of $\Delta MNP$ (the medial triangle of $\Delta ABC$) is $\frac{1}{4}$ of the area of $\Delta ABC$:
   \[
   [MNP] = \frac{1}{4}[ABC]
   \]

5. Substituting this into the previous equation, we get:
   \[
   [G_1G_2G_3] = \frac{4}{9}[MNP] = \frac{4}{9} \cdot \frac{1}{4}[ABC]
   \]

6. Simplifying the expression, we find:
   \[
   [G_1G_2G_3] = \frac{4}{36}[ABC] = \frac{1}{9}[ABC]
   \]

7. Given that the area of $\Delta G_1G_2G_3$ is 7 units, we can set up the equation:
   \[
   7 = \frac{1}{9}[ABC]
   \]

8. Solving for $[ABC]$, we multiply both sides by 9:
   \[
   [ABC] = 7 \times 9 = 63
   \]

The final answer is $\boxed{63}$ units.