In this paper we mainly employ the Zeilberger algorithm to study congruences for sums of terms involving products of three binomial coefficients. Let $p>3$ be a prime. We prove that $∑_{k=0}^{p-1}\frac{\binom{2k}k^2\binom{2k}{k+d}}{64^k}≡ 0\pmod{p^2}$ for all $d𝟄\{0,…,p-1\}$ with $d≡ (p+1)/2\pmod2$. If $p≡ 1\pmod4$ and $p=x^2+y^2$ with $x≡ 1\pmod4$ and $y≡ 0\pmod2$, then we show $∑_{k=0}^{p-1}\frac{\binom{2k}k^2\binom{2k}{k+1}}{(-8)^k}≡ 2p-2x^2\pmod{p^2}\ \ \mbox{and}\ \ ∑_{k=0}^{p-1}\frac{\binom{2k}k\binom{2k}{k+1}^2}{(-8)^k}≡-2p\pmod{p^2}$ by means of determining $x$ mod $p^2$ via $(-1)^{(p-1)/4}\,x≡∑_{k=0}^{(p-1)/2}\frac{k+1}{8^k}\binom{2k}k^2≡∑_{k=0}^{(p-1)/2}\frac{2k+1}{(-16)^k}\binom{2k}k^2\pmod{p^2}.$ We also solve the remaining open cases of Rodriguez-Villegas' conjectural congruences on $∑_{k=0}^{p-1}\frac{\binom{2k}k^2\binom{3k}k}{108^k},\ \ ∑_{k=0}^{p-1}\frac{\binom{2k}k^2\binom{4k}{2k}}{256^k}, \ \ ∑_{k=0}^{p-1}\frac{\binom{2k}{k}\binom{3k}k\binom{6k}{3k}}{12^{3k}}$ modulo $p^2$. ; Comment: 21 pages, final published version