Official solution
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
\[OG = \sqrt {{2^2} + {2^2}} = 2\sqrt 2 \]