Three identical spheres each of mass $$M$$ are placed at the corners of a right angled triangle with mutually perpendicular sides equal to $$3 \text{ m}$$ each. Taking point of intersection of mutually perpendicular sides as origin, the magnitude of position vector of centre of mass of the system will be $$\sqrt{x}$$ m. The value of $$x$$ is ______.

Three identical spheres, each of mass $$M$$, are placed at the corners of a right-angled triangle with mutually perpendicular sides of $$3 \text{ m}$$ each. Taking the point of intersection of the perpendicular sides as the origin, the three vertices are at:

$$A = (0, 0)$$, $$B = (3, 0)$$, $$C = (0, 3)$$

The coordinates of the centre of mass are:

$$x_{cm} = \dfrac{M(0) + M(3) + M(0)}{3M} = \dfrac{3}{3} = 1 \text{ m}$$

$$y_{cm} = \dfrac{M(0) + M(0) + M(3)}{3M} = \dfrac{3}{3} = 1 \text{ m}$$

The magnitude of the position vector of the centre of mass is:

$$r = \sqrt{x_{cm}^2 + y_{cm}^2} = \sqrt{1^2 + 1^2} = \sqrt{2} \text{ m}$$

Comparing with $$\sqrt{x}$$, we get $$x = 2$$.

Therefore, the answer is $$\boxed{2}$$.