H₂A is a weak diprotic acid. If the pH of 0.1M H₂A solution is 3 and concentration of A^2-  is 10^-12 at 298K. select correct statement(s)

$\begin{array}{l}{\mathrm{p}}^{\mathrm{H}}=\frac{1}{2}\left[{\mathrm{p}}^{{\mathrm{K}}_{{\mathrm{a}}_1}}-\log \mathrm{C}\right]\\ 3=\frac{1}{2}\left[{\mathrm{P}}^{{\mathrm{K}}_{{\mathrm{q}}_1}}+1\right];{\mathrm{p}}^{{\mathrm{K}}_{{\mathrm{a}}_1}}=5\Rightarrow {\mathrm{K}}_{{\mathrm{a}}_1}={10}^{-5}\\ {\mathrm{H}}_2\mathrm{A}\rightleftharpoons {\mathrm{H}}^{\oplus }+{\mathrm{HA}}^{-}\end{array}$
Option-a-Correct
At $\text{ Eq: }\mathrm{c}-\mathrm{x}\mathrm{x}+\mathrm{y}\approx \mathrm{x}\mathrm{x}-\mathrm{y}\approx \mathrm{x}$
y is very less. So $\mathrm{x}+\mathrm{y}\approx \mathrm{x}\mathrm{\&}\mathrm{x}-\mathrm{y}\approx \mathrm{x}$
$\begin{array}{l}\therefore {\mathrm{K}}_{{\mathrm{a}}_1}=\frac{{\mathrm{x}}^2}{\mathrm{c}-\mathrm{x}}\\ {\mathrm{HA}}^{-}\rightleftharpoons {\mathrm{H}}^{\oplus }+{\mathrm{A}}^{-}\end{array}$

At $\text{ eq: }\mathrm{x}-\mathrm{y}\approx \mathrm{x}\mathrm{y}\mathrm{y}$
$\begin{array}{l}{\mathrm{K}}_{{\mathrm{a}}_2}=\mathrm{y}={10}^{-12};{\mathrm{P}}^{{\mathrm{K}}_{{\mathrm{a}}_2}}=12\\ \therefore {\left[{\mathrm{H}}^{+}\right]}_{\text{total }}=\left[{\mathrm{H}}^{+}\right]\text{from }{\mathrm{H}}_2\mathrm{A}+\left[{\mathrm{H}}^{+}\right]\text{from }{\mathrm{HA}}^{-}\\ =\mathrm{x}+\mathrm{y}\approx \mathrm{x}\end{array}$
Option – b - Correct
From $\left(\mathrm{c}\right)\Rightarrow {\mathrm{P}}^{\mathrm{H}}=3\Rightarrow \left[{\mathrm{H}}^{+}\right]={10}^{-3}\mathrm{M}$         Incorrect
From $\left(\mathrm{d}\right)\Rightarrow {\mathrm{P}}^{{\mathrm{K}}_{{\mathrm{a}}_2}}-{\mathrm{P}}^{{\mathrm{K}}_{{\mathrm{a}}_1}}=12-5=7$           Incorrect