1. Find the vertices of the given triangle

The sides are on the lines $L_1: 4x-7y+10=0,$ $L_2: x+y=5,$ $L_3: 7x+4y=15.$

The vertices are the pairwise intersections.

• Vertex $A = L_2 \cap L_3$

  From $x+y=5$, we get $y=5-x$. Substituting into $7x+4y=15$: $7x+4(5-x)=15$ $7x+20-4x=15$ $3x=-5 \Rightarrow x=-\frac{5}{3}$ $y=5-\left(-\frac{5}{3}\right)=\frac{20}{3}$ So, $A\left(-\frac{5}{3},\frac{20}{3}\right).$

• Vertex $B = L_1 \cap L_3$

  Solve $4x-7y=-10,\qquad 7x+4y=15.$ Multiply first by $4$ and second by $7$: $16x-28y=-40,$ $49x+28y=105.$ Adding, $65x=65 \Rightarrow x=1.$ Then $7(1)+4y=15 \Rightarrow 4y=8 \Rightarrow y=2.$ So, $B(1,2).$

• Vertex $C = L_1 \cap L_2$

  From $x+y=5$, $y=5-x$. Substitute into $4x-7y+10=0$: $4x-7(5-x)+10=0$ $4x-35+7x+10=0$ $11x-25=0 \Rightarrow x=\frac{25}{11}$ $y=5-\frac{25}{11}=\frac{30}{11}.$ So, $C\left(\frac{25}{11},\frac{30}{11}\right).$

2. Find two altitudes of the triangle

The orthocentre is the intersection of altitudes.

Altitude through $B$

Side $AC$ lies on $L_2: x+y=5$, whose slope is $-1$. Hence an altitude perpendicular to it has slope $1$.

Through $B(1,2)$, the altitude is $y-2=1(x-1)$ $y=x+1.$

Altitude through $C$

Side $AB$ lies on $L_3: 7x+4y=15$. Its slope is $4y=-7x+15 \Rightarrow y=-\frac{7}{4}x+\frac{15}{4},$ so the perpendicular slope is $\frac{4}{7}$.

Through $C\left(\frac{25}{11},\frac{30}{11}\right)$, altitude is $y-\frac{30}{11}=\frac{4}{7}\left(x-\frac{25}{11}\right).$

3. Find the orthocentre of the given triangle

Use the altitude $y=x+1$ in the second altitude: $x+1-\frac{30}{11}=\frac{4}{7}\left(x-\frac{25}{11}\right).$

Left side: $x+1-\frac{30}{11}=x-\frac{19}{11}.$ So, $x-\frac{19}{11}=\frac{4}{7}\left(x-\frac{25}{11}\right).$ Multiply by $77$: $77x-133=44x-100$ $33x=33 \Rightarrow x=1.$ Then $y=x+1=2.$ Thus the orthocentre is $H=(1,2).$

4. Find the orthocentre of the triangle formed by $x=0$, $y=0$, $x+y=1$

These lines form the triangle with vertices $(0,0), (1,0), (0,1).$ This is a right triangle with right angle at $(0,0)$. Hence its orthocentre is the right-angled vertex itself: $H'=(0,0).$

5. Distance between the two orthocentres

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

So the required distance is $\boxed{\sqrt{5}}.$

6. Check with options

Option C is $\sqrt{5}.$ So the correct option is C.