1. Find the centroid of the second triangle

For any triangle, if $G$ is the centroid, $O$ the circumcentre, and $H$ the orthocentre, then

which gives the vector relation

or equivalently

Given:

So

Thus, the orthocentre of the triangle formed by the three given lines is $(0,0)$.

2. Given lines

The three sides of the triangle are:

Let their pairwise intersections be the vertices of the triangle.

We need the orthocentre of this triangle to be $(0,0)$.

3. Find two vertices explicitly

Let

• $A=L_2\cap L_3$
• $B=L_3\cap L_1$
• $C=L_1\cap L_2$

First find $C$ from

Subtracting $2(x+2y=1)$ from the first equation:

Then

So

4. Use altitude from $C$ passes through orthocentre $(0,0)$

Since orthocentre is $(0,0)$, the altitude from $C$ must pass through $(0,0)$.

Slope of altitude through $C=(-1,1)$ and $(0,0)$ is

Hence side $AB$ (which lies on $L_3$) must have slope perpendicular to this, i.e.

Now line $L_3:ax+by-1=0$ has slope

So

Thus $L_3$ is of the form

5. Use altitude from another vertex

Now find $A=L_2\cap L_3$.

Equations are

From the first,

Substitute into $a(x-y)=1$:

Then

So

Since orthocentre is $(0,0)$, altitude from $A$ passes through $(0,0)$. Therefore line $AO$ has slope

Hence side $BC$ (which lies on $L_1$) must have slope perpendicular to this.

Slope of $L_1:2x+3y-1=0$ is

So

Thus

Solve:

Therefore

6. Compute $|a-b|$

7. Compare with stored answer

Derived answer is

which matches the stored correct answer.