Balance the following equations in basic medium by ion-electron method and oxidation number methods and identify the oxidising agent and the reducing agent.
$(a)$ $P_{4(s)} + OH_{(aq)}^{-} \to PH_{3(g)} + HPO_{2(aq)}^{-}$
$(b)$ $N_{2}H_{4(l)} + ClO_{3(aq)}^{-} \to NO_{(g)} + Cl_{(g)}^{-}$
$(c)$ $Cl_{2}O_{7(g)} + H_{2}O_{2(aq)} \to ClO_{2(aq)}^{-} + O_{2(g)} + H^{+}$

(A) For reaction $(a)$: $P_{4(s)} + 3OH_{(aq)}^{-} + 3H_{2}O_{(l)} \to PH_{3(g)} + 3HPO_{2(aq)}^{-}$
$1$. Oxidation half-reaction: $P_{4} \to 4HPO_{2}^{-} + 8e^{-}$
$2$. Reduction half-reaction: $P_{4} + 12e^{-} \to 4PH_{3}$
$3$. Multiply oxidation half by $3$ and reduction half by $2$ to balance electrons: $3P_{4} + 2P_{4} + 12OH^{-} + 12H_{2}O \to 12HPO_{2}^{-} + 8PH_{3}$.
$4$. Simplified: $5P_{4} + 12OH^{-} + 12H_{2}O \to 8PH_{3} + 12HPO_{2}^{-}$.
Here,$P_{4}$ acts as both the oxidising and reducing agent (disproportionation reaction).