Use Kohlrausch’s law of independent migration of ions

At infinite dilution, $\Lambda_m^\infty(\text{electrolyte})=\lambda_+^\infty+\lambda_-^\infty$

We need: $\Lambda_{\mathrm{HOAc}}^\infty = \lambda_{\mathrm{H^+}}^\infty + \lambda_{\mathrm{OAc^-}}^\infty$

Write the given molar conductances

From the table: $\Lambda^\infty(\mathrm{HCl}) = \lambda_{\mathrm{H^+}}^\infty + \lambda_{\mathrm{Cl^-}}^\infty = 426.2$ $\Lambda^\infty(\mathrm{NaOAc}) = \lambda_{\mathrm{Na^+}}^\infty + \lambda_{\mathrm{OAc^-}}^\infty = 91.0$ $\Lambda^\infty(\mathrm{NaCl}) = \lambda_{\mathrm{Na^+}}^\infty + \lambda_{\mathrm{Cl^-}}^\infty = 126.5$

Combine these to get $\Lambda^\infty(\mathrm{HOAc})$

Add the first two equations and subtract the third: $\Lambda^\infty(\mathrm{HCl}) + \Lambda^\infty(\mathrm{NaOAc}) - \Lambda^\infty(\mathrm{NaCl})$

Substituting: $=(\lambda_{\mathrm{H^+}}^\infty + \lambda_{\mathrm{Cl^-}}^\infty) + (\lambda_{\mathrm{Na^+}}^\infty + \lambda_{\mathrm{OAc^-}}^\infty) - (\lambda_{\mathrm{Na^+}}^\infty + \lambda_{\mathrm{Cl^-}}^\infty)$

Cancelling common terms: $= \lambda_{\mathrm{H^+}}^\infty + \lambda_{\mathrm{OAc^-}}^\infty$

Hence, $\Lambda^\infty(\mathrm{HOAc}) = 426.2 + 91.0 - 126.5$ $= 517.2 - 126.5$ $= 390.7$

Match with options

$\boxed{\Lambda_{\mathrm{HOAc}}^\infty = 390.7}$

So, the correct option is C.