If the orthocentre of the triangle formed by the lines $$y = x + 1$$, $$y = 4x - 8$$ and $$y = mx + c$$ is at $$(3, -1)$$, then $$m - c$$ is :

The three sides of the triangle are
$$L_1: y = x + 1 \quad (m_1 = 1),$$
$$L_2: y = 4x - 8 \quad (m_2 = 4),$$
$$L_3: y = mx + c \quad (m_3 = m).$$

Their orthocentre is given to be $$H(3,-1).$$

Step 1: Find the vertex opposite to $$L_3$$.
Intersect $$L_1$$ and $$L_2$$:
$$x + 1 = 4x - 8 \;\Longrightarrow\; 3x = 9 \;\Longrightarrow\; x = 3,$$
$$y = 3 + 1 = 4.$$
Hence $$C(3,\,4).$$
Vertex $$C$$ is opposite the side $$L_3.$$

Step 2: Use the orthocentre to obtain the slope of $$L_3$$.
Altitude through $$C$$ must pass through $$H(3,-1).$$
Slope of $$CH$$ is
$$\frac{-1 - 4}{\,3 - 3\,} = \frac{-5}{0} \; \Longrightarrow\; \text{vertical line } x = 3.$$
Therefore $$L_3$$, being perpendicular to this altitude, must be horizontal:
$$m = 0 \; \Longrightarrow\; L_3: y = c.$$

Step 3: Express the remaining two vertices in terms of $$c$$.
Intersection of $$L_1$$ with $$L_3$$ gives
$$y = c = x + 1 \;\Longrightarrow\; B(c - 1,\,c).$$
Intersection of $$L_2$$ with $$L_3$$ gives
$$y = c = 4x - 8 \;\Longrightarrow\; x = \frac{c + 8}{4},$$
so $$A\left(\frac{c + 8}{4},\,c\right).$$

Step 4: Use the altitude from $$A$$ to determine $$c$$.
Slope of side $$BC$$ is
$$\frac{4 - c}{3 - (c - 1)} = \frac{4 - c}{4 - c} = 1.$$
Hence the altitude from $$A$$ is perpendicular to $$BC$$, so its slope is $$-1.$$
Equation of the altitude passing through the orthocentre $$H(3,-1)$$ is
$$y + 1 = -1\bigl(x - 3\bigr) \;\Longrightarrow\; y = -x + 2.$$ This altitude must also pass through $$A$$:
$$c = -\frac{c + 8}{4} + 2.$$ Multiply by 4:
$$4c = -(c + 8) + 8 = -c.$$ Therefore $$5c = 0 \;\Longrightarrow\; c = 0.$$

Step 5: Compute $$m - c$$.
We have $$m = 0$$ and $$c = 0,$$ hence
$$m - c = 0 - 0 = 0.$$

Thus $$m - c = 0.$$
Option A is correct.