Let $$A(a, b)$$, $$B(3, 4)$$ and $$(-6, -8)$$ respectively denote the centroid, circumcentre and orthocentre of a triangle. Then, the distance of the point $$P(2a + 3, 7b + 5)$$ from the line $$2x + 3y - 4 = 0$$ measured parallel to the line $$x - 2y - 1 = 0$$ is

The centroid $$G(a,b)$$, the circumcentre $$O(3,4)$$ and the orthocentre $$H(-6,-8)$$ of any triangle are collinear on the Euler line.
On this line the centroid divides the segment $$OH$$ in the ratio $$OG : GH = 1 : 2$$.

Using the section (internal-division) formula:
If a point $$P(x,y)$$ divides $$AB$$ internally in the ratio $$m:n$$, then $$x = \frac{mx_B + nx_A}{m+n}, \qquad y = \frac{my_B + ny_A}{m+n} \; -(1)$$

Here $$A \equiv O(3,4)$$, $$B \equiv H(-6,-8)$$ and $$P \equiv G(a,b)$$ with ratio $$m:n = 1:2$$ (since $$OG : GH = 1:2$$).
Applying $$(1)$$:

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

Therefore the centroid is $$G(0,0)$$.

The given point $$P(2a+3,\; 7b+5) = (2\cdot0+3,\; 7\cdot0+5) = (3,5).$$

We need the distance of $$P(3,5)$$ from the line $$L_1: 2x + 3y - 4 = 0$$ measured parallel to the line $$L_2: x - 2y - 1 = 0.$$

Step 1 - Direction along which to measure: The normal vector of $$L_2$$ is $$(1,-2)$$, so a direction vector parallel to $$L_2$$ is perpendicular to this, e.g. $$\mathbf{d} = (2,1)$$ (since $$(1,-2)\cdot(2,1)=0$$).

Step 2 - Equation of the line through $$P$$ in direction $$\mathbf{d}$$: $$x = 3 + 2t,\; y = 5 + t \quad (t \in \mathbb{R}).$$

Step 3 - Intersection of this line with $$L_1$$ (call the intersection $$Q$$).
Substitute $$x$$ and $$y$$ in $$2x + 3y - 4 = 0$$:

$$2(3+2t) + 3(5+t) - 4 = 0$$
$$6 + 4t + 15 + 3t - 4 = 0$$
$$17 + 7t = 0 \;\Longrightarrow\; t = -\frac{17}{7}.$$

Step 4 - Length of $$PQ$$ along the chosen direction:
Magnitude of direction vector $$\mathbf{d}$$ is $$|\mathbf{d}| = \sqrt{2^2 + 1^2} = \sqrt{5}.$$ Therefore

$$PQ = |t|\,|\mathbf{d}| = \frac{17}{7}\,\sqrt{5} = \frac{17\sqrt{5}}{7}.$$

Hence, the required distance is $$\frac{17\sqrt{5}}{7}$$.

Option C is correct.