Step 1: Kohlrausch's law says each ion contributes a fixed, characteristic amount to the total conductivity at infinite dilution, independent of the other ion it came with. Hence the limiting molar conductivity is just the sum of the cation and anion parts:
\[ \Lambda^\circ_m(\text{electrolyte}) = \lambda^\circ(\text{cation}) + \lambda^\circ(\text{anion}) \]
Step 2: Because ion contributions add, we can build the value for acetic acid from the ion contributions hidden inside the three given strong-electrolyte values.
Step 3: Break each into ions:
\[ \Lambda^\circ(HCl) = \lambda^\circ_{H^+} + \lambda^\circ_{Cl^-} \]
\[ \Lambda^\circ(NaAc) = \lambda^\circ_{Na^+} + \lambda^\circ_{Ac^-} \]
\[ \Lambda^\circ(NaCl) = \lambda^\circ_{Na^+} + \lambda^\circ_{Cl^-} \]
Step 4: Adding the first two and subtracting the third cancels \(Na^+\) and \(Cl^-\), leaving exactly \(\lambda^\circ_{H^+} + \lambda^\circ_{Ac^-}\), which is \(\Lambda^\circ(HAc)\):
\[ \Lambda^\circ(HAc) = \Lambda^\circ(HCl) + \Lambda^\circ(NaAc) - \Lambda^\circ(NaCl) \]
Step 5: Plug in \(425.9\), \(91.0\) and \(126.4\):
\[ \Lambda^\circ(HAc) = (425.9 + 91.0) - 126.4 = 516.9 - 126.4 \]
\[\boxed{\Lambda^\circ(HAc) = 390.5 \; S\,cm^2\,mol^{-1}}\]