If secθ - cosecθ = 0, then the value of $$2 \sin 2\theta ⁡-3 cos⁡2θ$$​ is: 

​$$\sec \theta - \cosec \theta = 0$$​​

​$$\begin{aligned}&\sec \theta = \frac{1}{\cos \theta}, \quad \cosec \theta = \frac{1}{\sin \theta} \\&sin 2\theta = 2 \sin \theta \cos \theta \\&\cos 2\theta = \cos^2 \theta - \sin^2 \theta \end{aligned}$$​​

From the given equation:

​$$\sec \theta - \csc \theta = 0$$​​

​$$\frac{1}{\cos \theta} = \frac{1}{\sin \theta}$$​​

Hence:

​$$\cos \theta = \sin \theta$$​​

Therefore, $$\theta = 45^0$$​​

since $$\cos 45^\circ = \sin 45^\circ$$​​

Now, we calculate

​$$\sin 2\theta - 3 \cos 2\theta$$​​

when  $$\theta = 45^\circ$$​​

Calculating for $$2\theta:$$​​

​$$2\theta = 90^\circ$$​​

Now, substitute into the trigonometric functions:

​$$\sin 2\theta = \sin 90^\circ = 1, \quad \cos 2\theta = \cos 90^\circ = 0$$​​

Using these values in the expression:

​$$2 \sin 2\theta - 3 \cos 2\theta = 2(1) - 3(0) = 2$$​​

Thus, the value of $$2 \sin 2\theta - 3 \cos 2\theta$$​ is 2.