1. Use the electronic configuration and number of unpaired electrons to infer colour
   In aqueous solution, the colour of transition metal ions mainly arises due to $d$-$d$ transitions. Ions with similar $d^n$ configurations often show similar colours, and ions with $d^0$ or $d^{10}$ are generally colourless.

2. Check each ion in the options

Option A: $\mathrm{V}^{2+}, \mathrm{Cr}^{3+}, \mathrm{Mn}^{3+}$

• $\mathrm{V}$: $[Ar]3d^34s^2$
  $\mathrm{V}^{2+} = [Ar]3d^3$

• $\mathrm{Cr}$: $[Ar]3d^54s^1$
  $\mathrm{Cr}^{3+} = [Ar]3d^3$

• $\mathrm{Mn}$: $[Ar]3d^54s^2$
  $\mathrm{Mn}^{3+} = [Ar]3d^4$

Known aqueous colours:

• $\mathrm{V}^{2+}$ : violet
• $\mathrm{Cr}^{3+}$ : violet (or green depending on complex/hydration, but commonly violet in aqueous hydrated form)
• $\mathrm{Mn}^{3+}$ : violet

So these are taken as same colour.

Option B: $\mathrm{Ti}^{4+}, \mathrm{V}^{4+}, \mathrm{Mn}^{2+}$

• $\mathrm{Ti}^{4+} = d^0$ → colourless
• $\mathrm{V}^{4+}$ (as vanadyl ion) → blue
• $\mathrm{Mn}^{2+} = d^5$ → pale pink

These are not same colour.

Option C: $\mathrm{Sc}^{3+}, \mathrm{Ti}^{3+}, \mathrm{Cr}^{2+}$

• $\mathrm{Sc}^{3+} = d^0$ → colourless
• $\mathrm{Ti}^{3+} = d^1$ → purple/violet
• $\mathrm{Cr}^{2+} = d^4$ → blue

These are not same colour.

Option D: $\mathrm{Zn}^{2+}, \mathrm{V}^{3+}, \mathrm{Fe}^{3+}$

• $\mathrm{Zn}^{2+} = d^{10}$ → colourless
• $\mathrm{V}^{3+} = d^2$ → green
• $\mathrm{Fe}^{3+} = d^5$ → yellow/brown

These are not same colour.

3. Conclusion

The only set considered to have the same colour in aqueous solution is:

$\boxed{\mathrm{V}^{2+},\ \mathrm{Cr}^{3+},\ \mathrm{Mn}^{3+}}$

So the correct option is A.