Let the two given sides of the triangle be

$L_1:x-2y+1=0$ $L_2:2x-y-1=0$

and the orthocenter be

$H\left(\frac73,\frac73\right).$

We need the distance of the origin from the centroid of the triangle.

1. Find the vertex formed by the two given sides

Since two sides of the triangle are given, their intersection gives one vertex, say $A$.

Solve

$x-2y+1=0 \implies x-2y=-1$ $2x-y-1=0 \implies 2x-y=1$

From the first equation,

$x=2y-1$

Substitute into the second:

$2(2y-1)-y=1$ $4y-2-y=1$ $3y=3 \implies y=1$

Hence

$x=2(1)-1=1$

So,

$A=(1,1).$

2. Use the orthocenter property to get the other two sides

If $A$ is a vertex and $H$ is the orthocenter, then the altitude from $B$ lies along $L_2$ and is perpendicular to side $AC$, while the altitude from $C$ lies along $L_1$ and is perpendicular to side $AB$.

So:

• side $AB$ passes through $A$ and is perpendicular to $L_1$
• side $AC$ passes through $A$ and is perpendicular to $L_2$

Slope of $L_1$

From

$x-2y+1=0 \implies y=\frac12 x+\frac12$

So slope of $L_1$ is

$m_1=\frac12.$

Therefore slope of $AB$ is

$m_{AB}=-2.$

Equation of $AB$ through $(1,1)$:

$y-1=-2(x-1)$ $y=-2x+3$ $2x+y-3=0.$

Slope of $L_2$

From

$2x-y-1=0 \implies y=2x-1$

So slope of $L_2$ is

$m_2=2.$

Therefore slope of $AC$ is

$m_{AC}=-\frac12.$

Equation of $AC$ through $(1,1)$:

$y-1=-\frac12(x-1)$ $2y-2=-x+1$ $x+2y-3=0.$

3. Find vertices $B$ and $C$

Vertex $B$

$B$ lies on side $AB$ and also on side $L_2$.

Solve

$2x+y-3=0$ $2x-y-1=0$

Add the equations:

$4x-4=0 \implies x=1$

Then from $2x-y-1=0$,

$2-y-1=0 \implies y=1$

This again gives $(1,1)$, so this labeling is not correct.

Let us assign carefully:

Since $L_1$ and $L_2$ are the two sides through $A$, they are actually $AB$ and $AC$ themselves.

Thus:

• $AB$ is $L_1:x-2y+1=0$
• $AC$ is $L_2:2x-y-1=0$

Then:

• altitude from $B$ passes through $H$ and is perpendicular to $AC$
• altitude from $C$ passes through $H$ and is perpendicular to $AB$

This is the correct approach.

4. Equation of altitude from $B$

Side $AC$ has slope $2$, so altitude from $B$ has slope

$-\frac12.$

Passing through

$H\left(\frac73,\frac73\right),$

its equation is

$y-\frac73=-\frac12\left(x-\frac73\right).$

Multiply by 6:

$6y-14=-3x+7$ $3x+6y-21=0$ $x+2y-7=0.$

So altitude from $B$ is

$x+2y-7=0.$

Since $B$ also lies on side $AB=L_1$,

$x-2y+1=0$ $x+2y-7=0$

Add:

$2x-6=0 \implies x=3$

Then

$3-2y+1=0 \implies 2y=4 \implies y=2.$

Hence

$B=(3,2).$

5. Equation of altitude from $C$

Side $AB$ has slope $\frac12$, so altitude from $C$ has slope

$-2.$

Passing through $H\left(\frac73,\frac73\right)$:

$y-\frac73=-2\left(x-\frac73\right).$

Multiply by 3:

$3y-7=-6x+14$ $6x+3y-21=0$ $2x+y-7=0.$

So altitude from $C$ is

$2x+y-7=0.$

Since $C$ lies on side $AC=L_2$,

$2x-y-1=0$ $2x+y-7=0$

Add:

$4x-8=0 \implies x=2$

Then

$4-y-1=0 \implies y=3.$

Hence

$C=(2,3).$

6. Find the centroid

The vertices are

$A=(1,1),\quad B=(3,2),\quad C=(2,3).$

So centroid $G$ is

$G\left(\frac{1+3+2}{3},\frac{1+2+3}{3}\right)=\left(2,2\right).$

7. Distance of origin from the centroid

Distance from $(0,0)$ to $(2,2)$ is

$\sqrt{2^2+2^2}=\sqrt{8}=2\sqrt2.$

8. Final answer

Thus the required distance is

$\boxed{2\sqrt2}.$

So the correct option is C.