Which one are correct about the solution that contains $3.42ppm$ $A{l}_2(S{O}_4{)}_3$ and $1.42ppm$ $N{a}_{2}S{O}_4$?

The correct options are
A $\left[A{l}^{3+}\right]=\left[N{a}^{+}\right]$
C $\left[S{O}_4^{2-}\right]=\left[N{a}^{+}\right]+\left[A{l}^{3+}\right]$
ppm = number of milligrams present in 1 litre of the solution

Given
Mass of $A{l}_2(S{O}_4{)}_3=3.42mg$
Mass of $N{a}_{2}S{O}_4=1.42mg$

$\left[A{l}_2\right(S{O}_4{)}_3]=\frac{3.42\times {10}^{-3}}{342}=0.01\times {10}^{-3}M$
$\left[S{O}_4^{2-}\right]=0.03\times {10}^{-3}M$
$\left[A{l}^{3+}\right]=0.02\times {10}^{-3}M$

$\left[N{a}_{2}S{O}_4\right]$$=\frac{1.42\times {10}^{-3}}{142}=0.01\times {10}^{-3}M=\left[S{O}_4^{2-}\right]$
$\left[N{a}^{+}\right]=0.02\times {10}^{-3}M$
total concentration of $\left[S{O}_4^{2-}\right]$ $=0.04\times {10}^{-3}M$
Hence in solution
$\left[N{a}^{+}\right]=0.02\times {10}^{-3}M$
$\left[S{O}_4^{2-}\right]$ $=0.04\times {10}^{-3}M$
$\left[A{l}^{3+}\right]=0.02\times {10}^{-3}M$
So,
$\left[A{l}^{3+}\right]=\left[N{a}^{+}\right]$ and $\left[S{O}_4^{2-}\right]=\left[N{a}^{+}\right]+\left[A{l}^{3+}\right]$