If the line $$x = y = z$$ intersects the line $$x\sin A + y\sin B + z\sin C - 18 = 0 = x\sin 2A + y\sin 2B + z\sin 2C - 9$$, where $$A$$, $$B$$, $$C$$ are the angles of a triangle $$ABC$$, then $$80\left(\sin\frac{A}{2}\sin\frac{B}{2}\sin\frac{C}{2}\right)$$ is equal to _____.

$$x = y = z = t$$

$$t(\sin A + \sin B + \sin C) = 18$$

$$t(\sin 2A + \sin 2B + \sin 2C) = 9$$

$$\frac{\sin 2A + \sin 2B + \sin 2C}{\sin A + \sin B + \sin C} = \frac{9}{18}$$

$$\frac{4 \sin A \sin B \sin C}{4 \cos \frac{A}{2} \cos \frac{B}{2} \cos \frac{C}{2}} = \frac{1}{2}$$

$$\frac{4 (2\sin\frac{A}{2}\cos\frac{A}{2}) (2\sin\frac{B}{2}\cos\frac{B}{2}) (2\sin\frac{C}{2}\cos\frac{C}{2})}{4 \cos \frac{A}{2} \cos \frac{B}{2} \cos \frac{C}{2}} = \frac{1}{2}$$

$$8 \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2} = \frac{1}{2}$$

$$\sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2} = \frac{1}{16}$$

$$80 \left(\sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2}\right) = 80 \left(\frac{1}{16}\right)$$

$$80 \left(\sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2}\right) = 5$$