Inclusion-exclusion is easiest when you first fix the "only" regions using totals and the given pairwise/three-way counts.
Step 1: Translate to "only" groups.
Given: \(|A|=15,\ |R|=12,\ |P|=11\).
A&P only \(=2\), A&R only \(=6\); R&P total \(=4\) includes all three \(=3\) \(\Rightarrow\) R&P only \(=1\).
Step 2: Find single-only counts.
A only \(= 15 - (6+2+3) = 4\).
R only \(= 12 - (6+1+3) = 2\).
P only \(= 11 - (2+1+3) = 5\).
Step 3: Sum "at least one".
Singles: \(4+2+5 = 11\).
Two-only: \(6+2+1 = 9\).
All three: \(3\).
Total with \(\ge 1\) option \(= 11+9+3 = 23\).
Step 4: Cars with none.
Total \(=25 \Rightarrow\) None \(= 25 - 23 = \boxed{2}\).
