Let the original speed of the train be $x$ km/hr.

The train travels 63 km at speed $x$ and 72 km at speed $x + 6$.

Time taken for each part of the journey: Time_1 = $63 / x$ Time_2 = $72 / (x + 6)$

Total time = 3 hours: $(63 / x) + (72 / (x + 6)) = 3$

Multiply the entire equation by $x(x + 6)$: $63(x + 6) + 72x = 3x(x + 6)$

Simplify: $63x + 378 + 72x = 3x^2 + 18x$ $135x + 378 = 3x^2 + 18x$ $3x^2 - 117x - 378 = 0$

Divide by 3: $x^2 - 39x - 126 = 0$

Solve using the quadratic formula: $x = [\frac{-(-39) \pm \sqrt{((-39)^2 - 4(1)(-126))}}{2(1)}]$ $x = [\frac{39 \pm \sqrt{1521 + 504}}{2}]$ $x = [\frac{39 \pm \sqrt{2025}}{2}]$ $x = [\frac{39 \pm 45}{2}]$

The two possible values for $x$: $x = (39 + 45) / 2 = 42$ or $x = (39 - 45) / 2 = -3$

Since speed cannot be negative, the original speed is 42 km/hr.