= Solution
Use the <orthonormality>[orthonormal] basis $|u\bar u\rangle,|d\bar d\rangle,|s\bar s\rangle$. The normalized <quark flavor>[flavour] <singlet representation>[singlet] is
$$
\boxed{|1\rangle=\frac{|u\bar u\rangle+|d\bar d\rangle+|s\bar s\rangle}{\sqrt3},
\qquad I=I_3=S=0.}
$$
The <SU(3)> invariant contraction pairs each fundamental <quark flavor>[flavour] with its conjugate. Applying any supplied raising operator to the <singlet representation>[singlet] gives zero because its <quark> and <antiquark> contributions cancel.
To fix the neutral <isospin> <triplet representation>[triplet] phase consistently with the source's charged states, apply $I_+$ to $-|d\bar u\rangle$:
$$
I_+[-|d\bar u\rangle]=|d\bar d\rangle-|u\bar u\rangle.
$$
Thus one missing member of the <meson octet> is
$$
\boxed{|T_0\rangle=\frac{|d\bar d\rangle-|u\bar u\rangle}{\sqrt2},\qquad I=1,\ I_3=S=0.}
$$
Indeed $I_+|T_0\rangle=\sqrt2|u\bar d\rangle$, the correct <ladder operator> strength for the $I=1$ <triplet representation>[triplet].
The other missing state is orthogonal to both $|1\rangle$ and $|T_0\rangle$. For coefficients $(a,b,c)$, <orthogonality> requires $a=b$ and $a+b+c=0$. Choose the convenient overall phase $c>0$ and normalize:
$$
\boxed{|8_0\rangle=\frac{2|s\bar s\rangle-|u\bar u\rangle-|d\bar d\rangle}{\sqrt6},
\qquad I=I_3=S=0.}
$$
It can also be obtained by adding the supplied ladder operations $V_+[-|s\bar u\rangle]=|s\bar s\rangle-|u\bar u\rangle$ and $U_+[-|s\bar d\rangle]=|s\bar s\rangle-|d\bar d\rangle$. The three displayed states are mutually <orthonormality>[orthonormal]. The <meson octet> here is the traceless adjoint component of $\bar3\otimes3$; the source's word “antisymmetric” does not impose antisymmetry on all its quark-antiquark <quark flavor>[flavour] wavefunctions.
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