Let the three sides of a triangle are on the lines $$4x - 7y + 10 = 0$$, $$x + y = 5$$ and $$7x + 4y = 15$$. Then the distance of its orthocentre from the orthocentre of the triangle formed by the lines $$x = 0$$, $$y = 0$$ and $$x + y = 1$$ is

The given three sides are the lines $$4x - 7y + 10 = 0$$, $$x + y = 5$$ and $$7x + 4y = 15$$.
Label them as
$$L_1 : 4x - 7y + 10 = 0$$,  $$L_2 : x + y - 5 = 0$$,  $$L_3 : 7x + 4y - 15 = 0$$.

Step 1  -  Find the three vertices.

Intersection of $$L_1$$ and $$L_2$$ :
Solve $$4x - 7y + 10 = 0$$ $$x + y = 5$$ From the second, $$y = 5 - x$$. Substitute: $$4x - 7(5 - x) + 10 = 0$$ $$\Rightarrow 4x - 35 + 7x + 10 = 0$$ $$\Rightarrow 11x = 25$$ $$\Rightarrow x = \frac{25}{11}, \; y = 5 - \frac{25}{11} = \frac{30}{11}.$$ Thus $$A\left(\frac{25}{11},\; \frac{30}{11}\right).$$

Intersection of $$L_2$$ and $$L_3$$ :
Solve $$x + y = 5$$ $$7x + 4y = 15$$ Put $$y = 5 - x$$ in the second: $$7x + 4(5 - x) = 15$$ $$\Rightarrow 7x + 20 - 4x = 15$$ $$\Rightarrow 3x = -5$$ $$\Rightarrow x = -\frac{5}{3},\; y = 5 - \left(-\frac{5}{3}\right) = \frac{20}{3}.$$ Thus $$B\left(-\frac{5}{3},\; \frac{20}{3}\right).$$

Intersection of $$L_1$$ and $$L_3$$ :
Solve $$4x - 7y + 10 = 0$$ $$7x + 4y = 15$$ From the second, $$y = \frac{15 - 7x}{4}.$$ Substitute in the first: $$4x - 7\left(\frac{15 - 7x}{4}\right) + 10 = 0$$ Multiply by $$4$$: $$16x - 7(15 - 7x) + 40 = 0$$ $$\Rightarrow 16x - 105 + 49x + 40 = 0$$ $$\Rightarrow 65x = 65$$ $$\Rightarrow x = 1,\; y = \frac{15 - 7(1)}{4} = 2.$$ Thus $$C(1,\,2).$$

Step 2  -  Locate the right angle of the triangle.

Slope of $$L_3 \;(BC)$$: $$7x + 4y = 15 \;\Rightarrow\; y = -\frac{7}{4}x + \frac{15}{4},$$ so $$m_{BC} = -\frac{7}{4}.$$

Slope of $$L_1 \;(AC)$$: $$4x - 7y + 10 = 0 \;\Rightarrow\; y = \frac{4}{7}x + \frac{10}{7},$$ so $$m_{AC} = \frac{4}{7}.$$

Product of slopes: $$m_{BC}\, m_{AC} = \left(-\frac{7}{4}\right)\left(\frac{4}{7}\right) = -1.$$ Hence $$AC \perp BC,$$ so the triangle is right-angled at the common vertex $$C(1,2).$$

Step 3  -  Orthocentre of the first triangle.

In a right triangle, the orthocentre coincides with the right-angle vertex. Therefore $$H_1 = C(1,\,2).$$

Step 4  -  Orthocentre of the second triangle.

The second triangle is bounded by $$x = 0$$, $$y = 0$$ and $$x + y = 1$$. Its vertices are $$(0,0),\; (1,0),\; (0,1),$$ with the right angle at the origin. Hence its orthocentre is $$H_2 = (0,\,0).$$

Step 5  -  Distance between the two orthocentres.

$$\text{Distance} = \sqrt{(1 - 0)^2 + (2 - 0)^2} = \sqrt{1 + 4} = \sqrt{5}.$$

Answer : $$\sqrt{5}$$  (Option B)