For point A,

2x $$-$$ y $$-$$ 1 = 0 ...... (1)

x $$-$$ 2y + 1 = 0 ...... (2)

Solving (1) and (2), we get

x = 1, y = 1.

$$\therefore$$ Point A = (1, 1)

Altitude from B to line AC is perpendicular to line AC.

$$\therefore$$ Equator of altitude BH is

2x + y + $$\lambda$$ = 0 ...... (3)

It passes through point $$H\left( {{7 \over 3},{7 \over 3}} \right)$$ so it satisfy the equation (3).

$${{14} \over 3} + {7 \over 3} + \lambda = 0$$

$$\Rightarrow$$ $$\alpha$$ = $$-$$7

$$\therefore$$ Altitude BH = 2x + y $$-$$ 7 = 0 ...... (4)

Solving equation (1) and (4), we get

x = 2, y = 3.

$$\therefore$$ Point B = (2, 3)

Altitude from C to line AB is perpendicular to line AB.

$$\therefore$$ Equation of altitude CH is

x + 2y + $$\lambda$$ = 0 ...... (5)

It passes through point $$H\left( {{7 \over 3},{7 \over 3}} \right)$$ so it satisfy equation (5).

$${7 \over 3} + {{14} \over 3} + \lambda = 0$$

$$\Rightarrow$$ $$\lambda$$ = $$-$$7

$$\therefore$$ Altitude CH = x + 2y $$-$$ 7 = 0 ...... (6)

Solving equation (2) and (6), we get

x = 3, y = 2

$$\therefore$$ Point C = (3, 2)

Centroid G (x, y) of triangle A (1, 1), B (2, 3) and C (3, 2) is

$$x = {{1 + 2 + 3} \over 3} = 2$$, $$y = {{1 + 2 + 3} \over 3} = 2$$

Now, distance of point G (2, 2) from center O (0, 0) is

$$OG = \sqrt {{2^2} + {2^2}} = 2\sqrt 2 $$