$LC$ in parallel with $R$ circucit resonance

I’ve been doing an experiment which involves measuring the voltage across the capacitor at different driving frequencies.
Consider the circuit below:

A sketch of the experimental curve $(R_1<R_2<R_3)$ with $(L,C)$ constant:

However I am still trying to understand why resonance takes place in the first place.
My logic is as follows:

$V_L$ and $V_C$ are in antiphase with each other (the inductor is in phase with the capacitor), so the voltage of the LC combination and hence the voltage of the circuit is:
$$V=V_L-V_C$$
$V$ is in phase with $I_R$.
The circuit current is
$$I=sqrt{I_{LC}^2+I_R^2}$$
$$frac{V}{Z}=sqrt{left(frac{V}{Z_{LC}}right)^2+left(frac{V}{R}right)^2}$$.
Since $V=V_{LC}=V_L-V-C$, $I_{LC}Z_{LC}=I_{LC}X_L-I_{LC}X_C$,
$$Z_{LC}=X_L-X_{C}$$
$$frac{1}{Z}=sqrt{frac{1}{(X_L-X_C)^2}+frac{1}{R^2}} to Z=frac{|X_L-X_R|R}{R^2+(X_L-X_C)^2}$$
For resonance, $X_L=X_C to Z=0$.
But $Z=0$ would lead to $I to infty$ so something isn’t right with my derivation for $Z$ (circuit impedence)?

Also, I have been given the amplitude of the resonance peak:
$$A=frac{k}{sqrt{(omega^2-omega_0^2)^2+(gamma omega)^2}}$$
where $k$ is some normalisation factor, $omega=2pi f$ and $gamma$ is the damping coefficient.
For $A$ to increase with $R$, am I correct to assume that $gamma propto frac{1}{R}$? Is there a differential equation to describe the circuit?

EDIT:
The voltage source has an internal reactance of $600 Omega$, $R_1,R_2,R_3=1,2,3 Omega$
I obtained the folliwing differential eq.:
$$ddot{I}_{LC}(t)+frac{1}{L}frac{R_{output}R}{R+R_{output}}dot{I}_{LC(t)}+frac{I_{LC}(t)}{LC}=frac{RV_0 omega cos(omega t)}{(R+R_{output})L}$$, where $omega$ is the driving angular frequency and $V_0$ is the peak driving voltage. This means $gamma=frac{R_{output}R}{L(R+R_{output})}$ even if the output resistance is taken into account, but wouldn’t higher R lead to higher $gamma$ and hence lower amplitude?($R_{output}=text{internal r in the source voltage}$)