If the orthocentre of the triangle formed by the lines $$2x + 3y - 1 = 0$$, $$x + 2y - 1 = 0$$ and $$ax + by - 1 = 0$$, is the centroid of another triangle, whose circumcentre and orthocentre respectively are $$(3, 4)$$ and $$(-6, -8)$$, then the value of $$|a - b|$$ is ________

The orthocentre of the triangle formed by the lines $$2x + 3y - 1 = 0$$, $$x + 2y - 1 = 0$$, and $$ax + by - 1 = 0$$ is given to be the centroid of another triangle. For this other triangle, the circumcentre is $$(3, 4)$$ and the orthocentre is $$(-6, -8)$$.

First, recall that in any triangle, the centroid $$G$$, circumcentre $$O$$, and orthocentre $$H$$ lie on the Euler line, and the centroid divides the segment joining the orthocentre and circumcentre in the ratio $$2:1$$, with the circumcentre closer to the centroid. The position vector of the centroid is given by:

$$\vec{G} = \frac{\vec{H} + 2\vec{O}}{3}$$

Substituting the given points $$O' = (3, 4)$$ and $$H' = (-6, -8)$$:

$$G' = \left( \frac{-6 + 2 \times 3}{3}, \frac{-8 + 2 \times 4}{3} \right) = \left( \frac{-6 + 6}{3}, \frac{-8 + 8}{3} \right) = \left( \frac{0}{3}, \frac{0}{3} \right) = (0, 0)$$

Thus, the centroid $$G'$$ is $$(0, 0)$$. This centroid is also the orthocentre of the triangle formed by the lines $$2x + 3y - 1 = 0$$, $$x + 2y - 1 = 0$$, and $$ax + by - 1 = 0$$. Therefore, the orthocentre of this triangle is $$(0, 0)$$.

To find the orthocentre of the triangle formed by the three lines, denote the lines as:

L1: $$2x + 3y = 1$$

L2: $$x + 2y = 1$$

L3: $$ax + by = 1$$

The orthocentre $$(0, 0)$$ satisfies the condition that the line from the orthocentre to each vertex is perpendicular to the opposite side. The vertices of the triangle are the pairwise intersections of these lines.

Find the vertices:

1. Intersection of L1 and L2 (vertex A):

Solve $$2x + 3y = 1$$ and $$x + 2y = 1$$.

From L2: $$x = 1 - 2y$$

Substitute into L1: $$2(1 - 2y) + 3y = 1 \implies 2 - 4y + 3y = 1 \implies 2 - y = 1 \implies y = 1$$

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

So, vertex A: $$(-1, 1)$$

2. Intersection of L1 and L3 (vertex B):

Solve $$2x + 3y = 1$$ and $$ax + by = 1$$.

Using elimination:

Multiply first equation by $$b$$: $$2b x + 3b y = b$$

Multiply second equation by $$3$$: $$3a x + 3b y = 3$$

Subtract: $$(2b x + 3b y) - (3a x + 3b y) = b - 3 \implies (2b - 3a)x = b - 3$$

So, $$x = \frac{b - 3}{2b - 3a}$$

From first equation multiplied by $$a$$: $$2a x + 3a y = a$$

Second equation multiplied by $$2$$: $$2a x + 2b y = 2$$

Subtract: $$(2a x + 3a y) - (2a x + 2b y) = a - 2 \implies (3a - 2b)y = a - 2$$

So, $$y = \frac{a - 2}{3a - 2b} = -\frac{a - 2}{2b - 3a}$$ (since $$3a - 2b = -(2b - 3a)$$)

Thus, vertex B: $$\left( \frac{b - 3}{2b - 3a}, -\frac{a - 2}{2b - 3a} \right)$$

3. Intersection of L2 and L3 (vertex C):

Solve $$x + 2y = 1$$ and $$ax + by = 1$$.

From L2: $$x = 1 - 2y$$

Substitute into L3: $$a(1 - 2y) + b y = 1 \implies a - 2a y + b y = 1 \implies (b - 2a)y = 1 - a$$

So, $$y = \frac{1 - a}{b - 2a}$$

Then $$x = 1 - 2y = 1 - 2\frac{1 - a}{b - 2a} = \frac{b - 2a - 2(1 - a)}{b - 2a} = \frac{b - 2a - 2 + 2a}{b - 2a} = \frac{b - 2}{b - 2a}$$

Thus, vertex C: $$\left( \frac{b - 2}{b - 2a}, \frac{1 - a}{b - 2a} \right)$$

Since the orthocentre is $$(0, 0)$$, the vector from $$(0, 0)$$ to each vertex is perpendicular to the direction vector of the opposite side.

For vertex A$$(-1, 1)$$, the opposite side is BC. Since B and C lie on L3, BC is the line L3: $$ax + by = 1$$. The normal vector to L3 is $$(a, b)$$, so the direction vector of BC is $$(-b, a)$$. The vector OA is $$(-1, 1)$$. Since AH is perpendicular to BC:

$$(-1, 1) \cdot (-b, a) = 0 \implies (-1)(-b) + (1)(a) = b + a = 0 \implies a + b = 0 \quad \text{(Equation 1)}$$

For vertex B$$\left( \frac{b - 3}{2b - 3a}, -\frac{a - 2}{2b - 3a} \right)$$, the opposite side is AC. Since A and C lie on L2, AC is the line L2: $$x + 2y = 1$$. The normal vector to L2 is $$(1, 2)$$, so the direction vector of AC is $$(-2, 1)$$. The vector OB is $$\left( \frac{b - 3}{2b - 3a}, -\frac{a - 2}{2b - 3a} \right)$$. Since BH is perpendicular to AC:

$$\left( \frac{b - 3}{2b - 3a} \right)(-2) + \left( -\frac{a - 2}{2b - 3a} \right)(1) = 0$$

Since the denominator $$2b - 3a \neq 0$$ (assuming a non-degenerate triangle), multiply by $$2b - 3a$$:

$$-2(b - 3) - (a - 2) = 0 \implies -2b + 6 - a + 2 = 0 \implies -a - 2b + 8 = 0 \implies a + 2b = 8 \quad \text{(Equation 2)}$$

Solving Equations 1 and 2:

From Equation 1: $$a + b = 0 \implies a = -b$$

Substitute into Equation 2: $$-b + 2b = 8 \implies b = 8$$

Then $$a = -b = -8$$

So, $$a = -8$$, $$b = 8$$

Now, verify for vertex C$$\left( \frac{b - 2}{b - 2a}, \frac{1 - a}{b - 2a} \right)$$. With $$a = -8$$, $$b = 8$$:

$$x_c = \frac{8 - 2}{8 - 2(-8)} = \frac{6}{8 + 16} = \frac{6}{24} = \frac{1}{4}$$

$$y_c = \frac{1 - (-8)}{8 - 2(-8)} = \frac{9}{24} = \frac{3}{8}$$

So, vertex C: $$\left( \frac{1}{4}, \frac{3}{8} \right)$$

The opposite side is AB. Since A and B lie on L1, AB is the line L1: $$2x + 3y = 1$$. The normal vector to L1 is $$(2, 3)$$, so the direction vector of AB is $$(-3, 2)$$. The vector OC is $$\left( \frac{1}{4}, \frac{3}{8} \right)$$. Since CH is perpendicular to AB:

$$\left( \frac{1}{4} \right)(-3) + \left( \frac{3}{8} \right)(2) = -\frac{3}{4} + \frac{6}{8} = -\frac{3}{4} + \frac{3}{4} = 0$$

Thus, the condition is satisfied.

Check the denominators for non-zero values:

$$2b - 3a = 2(8) - 3(-8) = 16 + 24 = 40 \neq 0$$

$$b - 2a = 8 - 2(-8) = 8 + 16 = 24 \neq 0$$

The slopes of the lines are distinct: L1 has slope $$-\frac{2}{3}$$, L2 has slope $$-\frac{1}{2}$$, L3 (with $$a = -8$$, $$b = 8$$) is $$-8x + 8y = 1$$, so slope $$\frac{8}{8} = 1$$. Thus, no parallel lines, and a triangle is formed.

Now, compute $$|a - b|$$:

$$|a - b| = |-8 - 8| = |-16| = 16$$

Therefore, the value of $$|a - b|$$ is 16.