For any odd prime p we obtain q-analogues of Van Hamme's supercongruence: $$ \sum_{ k=0}^{ \frac{ p-1}{ 2}}{ 2k\choose k}^3\frac{ 1}{ 64^k} \equiv 0 \pmod{ p^2} \quad\text{ for}\quad p\equiv 3\pmod 4, $$ and Rodriguez-Villegas' Beukers-like supercongruences involving products of three binomial coefficients. For example, we prove that \begin{ align*} \sum_{ k=0}^{ \frac{ p-1}{ 2}} { 2k\brack k}_{ q^2}^3 \frac{ q^{ 2k}}{ (-q^2;q^2)_k^2 (-q;q)_{ 2k}^2} &\equiv 0\pmod{ [p]^2} \quad\text{ for}\quad p\equiv 3\pmod 4, \\ \sum_{ k=0}^{ \frac{ p-1}{ 2}}{ 2k\brack k}_{ q^3}\frac{ (q;q^3)_k (q^{ 2};q^3)_{ k} q^{ 3k} }{ (q^{ 6};q^{ 6})_k^2 } &\equiv 0 \pmod{ [p]^2}\quad\text{ for}\quad p\equiv 2\pmod{ 3}, \end{ align*} where $[p]=1+q+\cdots+q^{ p-1}$, $(a;q)_n=(1-a)(1-aq)\cdots(1-aq^{ n-1})$, and ${ n\brack k}_q$ denotes the q-binomial coefficient. Actually, our results give q-analogues of Z.-H. Sun's and Z.-W. Sun's generalizations of the above Beukers-like supercongruences. Our proof uses the theory of basic hypergeometric series including a new q-Clausen-type summation formula.
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