Why is particular solution zero for an impulse excitation signal?
The correct form of the statement should be

the particular solution is zero for $t&gt;0$

This is simply the case because the input $\delta(t)$ is zero for $t&gt;0$.
So for $t&gt;0$, the impulse response equals the homogeneous solution. For $t&lt;0$, the impulse response is zero because the system is causal. What remains is the moment $t=0$. The only thing that can happen there is an impulse (or its derivatives but let's not get into that here). So the complete form of the impulse response is
$$h(t)=A_0\delta(t)+y_h(t)\tag{1}$$
where $y_h(t)$ is the homogeneous solution of the differential equation. The constant $A_0$ can be determined by looking at the highest derivative of the input on both sides of the differential equation. For the given system we have
$$h'(t)+\frac{h(t)}{RC}=\frac{\delta(t)}{RC}\tag{2}$$
With $h(t)$ given by $(1)$ we see that on the left-hand side of $(2)$ we have a term $A_0\delta'(t)$ with nothing to match on the right-hand side. Consequently, for this specific differential equation we obtain $A_0=0$.
Note that it's easy to find a system for which $A_0\neq 0$. E.g., for a first-order RC highpass filter we have the following differential equation for the impulse response:
$$h'(t)+\frac{h(t)}{RC}=\delta'(t)\tag{3}$$
Using $(1)$ and matching the terms with $\delta'(t)$ on both sides of $(3)$ we obtain $A_0=1$, resulting in
$$h(t)=\delta(t)-\frac{1}{RC}e^{-t/RC},\qquad t\ge 0\tag{4}$$