Balancing Redox acid?

The reaction occurs in a slightly acidic or neutral medium. In an alkaline medium, the product Cr₂O₇²⁻ would convert to CrO₄²⁻.
Cr₂O₇²⁻(aq) + 2OH⁻(aq) → 2CrO₄²⁻(aq) + H₂O(l)

Reduction half equation: MnO₄⁻(aq) + 4H⁺(aq) + 3e⁻ → MnO₂(s) + 2H₂O(l) ……. [1]
Oxidation half equation: 2Cr³⁺(aq) + 7H₂O(l) → Cr₂O₇²⁻(aq) + 14H⁺ + 6e⁻ …… [2]

[1]×2 + [2], and cancel 6e⁻, 8H⁺(aq) and 4H₂O(l) on the both sides
2MnO₄⁻(aq) + 2Cr³⁺(aq) + 3H₂O(l) → Cr₂O₇²⁻(aq) + 2MnO₂(s) + 6H⁺(aq)

2 MnO4{-} + 2 Cr{3+} + 6 OH{-} → Cr2O7{2-} + 2 MnO2 + 3 H2O