At least two properties

Sum pairwise. $20+15+10=45$. But items with all three are counted $3$ times, we want once.

Correct. Items with exactly two: pairwise sum $-3\cdot$(triple)? No — items with all three contribute $3$ to pairwise sum (each pair) but should contribute only once. Items with exactly two contribute $1$ to pairwise sum. So: at least two = (pairwise sum) $-2\cdot$(triple) + (triple)? This is getting confused.

Correct approach. At least two = (exactly 2) + (exactly 3). Let $t_3=|A\cap B\cap C|=5$, $t_2=|$exactly two$|$. Pairwise sum = $2t_2+3t_3$ if we count per item? No: pairwise sum counts each item with exactly two properties once, and each item with three properties $3$ times. So pairwise sum $=t_2+3t_3=45$, giving $t_2=45-15=30$. At least two $=t_2+t_3=35$.

Conclusion. $35$ items have at least two properties.