The measures, in degrees, of the angles , $\alpha, \beta$ and $\theta$ are greater than $0$ less than $60$. Find the value of $\theta$ knowing, also, that $\alpha + \beta = 2\theta$ and that $$\sin \alpha \sin \beta \sin \theta = \sin(60 - \alpha ) \sin(60 - \beta) \sin(60 - \theta ).$$
Given the conditions:
1. \( \alpha + \beta = 2\theta \)
2. \( \sin \alpha \sin \beta \sin \theta = \sin(60^\circ - \alpha) \sin(60^\circ - \beta) \sin(60^\circ - \theta) \)

We need to find the value of \( \theta \).

Let's start by analyzing the given trigonometric identity.

1. **Rewrite the given identity:**
   \[
   \sin \alpha \sin \beta \sin \theta = \sin(60^\circ - \alpha) \sin(60^\circ - \beta) \sin(60^\circ - \theta)
   \]

2. **Use the product-to-sum identities:**
   \[
   \sin x \sin y = \frac{1}{2} [\cos(x-y) - \cos(x+y)]
   \]
   Applying this to both sides of the equation, we get:
   \[
   \sin \alpha \sin \beta = \frac{1}{2} [\cos(\alpha - \beta) - \cos(\alpha + \beta)]
   \]
   and
   \[
   \sin(60^\circ - \alpha) \sin(60^\circ - \beta) = \frac{1}{2} [\cos((60^\circ - \alpha) - (60^\circ - \beta)) - \cos((60^\circ - \alpha) + (60^\circ - \beta))]
   \]

3. **Simplify the expressions:**
   \[
   \sin \alpha \sin \beta = \frac{1}{2} [\cos(\alpha - \beta) - \cos(2\theta)]
   \]
   and
   \[
   \sin(60^\circ - \alpha) \sin(60^\circ - \beta) = \frac{1}{2} [\cos(\alpha - \beta) - \cos(120^\circ - 2\theta)]
   \]

4. **Substitute these into the original equation:**
   \[
   \frac{1}{2} [\cos(\alpha - \beta) - \cos(2\theta)] \sin \theta = \frac{1}{2} [\cos(\alpha - \beta) - \cos(120^\circ - 2\theta)] \sin(60^\circ - \theta)
   \]

5. **Simplify the equation:**
   \[
   [\cos(\alpha - \beta) - \cos(2\theta)] \sin \theta = [\cos(\alpha - \beta) - \cos(120^\circ - 2\theta)] \sin(60^\circ - \theta)
   \]

6. **Factor out \(\cos(\alpha - \beta)\):**
   \[
   \cos(\alpha - \beta) [\sin \theta - \sin(60^\circ - \theta)] = \cos(120^\circ - 2\theta) \sin(60^\circ - \theta) - \cos(2\theta) \sin \theta
   \]

7. **Use angle sum and difference identities:**
   \[
   \sin(60^\circ - \theta) = \sin 60^\circ \cos \theta - \cos 60^\circ \sin \theta = \frac{\sqrt{3}}{2} \cos \theta - \frac{1}{2} \sin \theta
   \]

8. **Substitute and simplify:**
   \[
   \cos(\alpha - \beta) \left[\sin \theta - \left(\frac{\sqrt{3}}{2} \cos \theta - \frac{1}{2} \sin \theta\right)\right] = \cos(120^\circ - 2\theta) \left(\frac{\sqrt{3}}{2} \cos \theta - \frac{1}{2} \sin \theta\right) - \cos(2\theta) \sin \theta
   \]

9. **Simplify further:**
   \[
   \cos(\alpha - \beta) \left[\frac{3}{2} \sin \theta - \frac{\sqrt{3}}{2} \cos \theta\right] = \cos(120^\circ - 2\theta) \left(\frac{\sqrt{3}}{2} \cos \theta - \frac{1}{2} \sin \theta\right) - \cos(2\theta) \sin \theta
   \]

10. **Assume \(\theta = 30^\circ\):**
    \[
    \alpha + \beta = 2 \times 30^\circ = 60^\circ
    \]
    \[
    \sin 30^\circ = \frac{1}{2}
    \]
    \[
    \sin(60^\circ - 30^\circ) = \sin 30^\circ = \frac{1}{2}
    \]

11. **Verify the solution:**
    \[
    \sin \alpha \sin \beta \sin 30^\circ = \sin(60^\circ - \alpha) \sin(60^\circ - \beta) \sin 30^\circ
    \]
    \[
    \sin \alpha \sin \beta \frac{1}{2} = \sin(60^\circ - \alpha) \sin(60^\circ - \beta) \frac{1}{2}
    \]
    \[
    \sin \alpha \sin \beta = \sin(60^\circ - \alpha) \sin(60^\circ - \beta)
    \]

Since the equation holds true, the value of \(\theta\) is indeed \(30^\circ\).

The final answer is \(\boxed{30^\circ}\).