= Solution
The original diagram has a <series circuit> containing $R$ and $L$, in parallel with $C$; it is not three separate parallel components. The <electrical admittance> is therefore
$$
Y=\frac1Z=\frac1{R+i\omega L}+i\omega C=\frac R{R^2+\omega^2L^2}+i\left[\omega C-\frac{\omega L}{R^2+\omega^2L^2}\right].
$$
For $R>0$, voltage and current are in phase when the <imaginary part> of $Y$ vanishes. Excluding the zero-<frequency> solution, division by $\omega$ gives
$$
C=\frac L{R^2+\omega_r^2L^2},\qquad\boxed{\omega_r=\sqrt{\frac1{LC}-\frac{R^2}{L^2}}.}
$$
A nonzero <lossy parallel coil resonance> exists only if $L>R^2C$. With $R=10\,\Omega$, $L=0.150\,\mathrm H$ and $C=1.00\times10^{-5}\,\mathrm F$,
$$
\omega_r=\sqrt{666666.7-4444.44}\,\mathrm{s^{-1}}=\boxed{813.77\,\mathrm{rad\,s^{-1}}},\qquad\boxed{f_r=\frac{\omega_r}{2\pi}=129.52\,\mathrm{Hz}.}
$$
For zero <electrical resistance>, the ideal parallel <inductor>-<capacitor> cancellation occurs at
$$
\boxed{\omega_0=\frac1{\sqrt{LC}}=816.50\,\mathrm{rad\,s^{-1}},\qquad f_0=129.95\,\mathrm{Hz}.}
$$
The fractional reduction relative to the zero-resistance coil is
$$
\boxed{\frac{f_0-f_r}{f_0}=1-\sqrt{1-\frac{R^2C}{L}}=0.00333891=0.3339\%.}
$$
At exactly $R=0$ and $\omega=\omega_0$, $Y=0$, so finite applied voltage draws zero net current in the ideal model. The quoted <frequency> is the limiting parallel-resonance value; the phase of an exactly zero current is undefined.
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