2 elaboration of why one might use the second formulation.
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The alternative hypothesis is β1≠0 OR β2≠0 OR β3≠0, via De Morgan's Laws. Proving any of those three conditions would disprove the null hypothesis.

The breakdown β1≠β2 OR β2≠β3 OR β3≠0 is mathematically equivalent but might be easier to show significance or design an experiment for.

The alternative hypothesis is β1≠0 OR β2≠0 OR β3≠0, via De Morgan's Laws. Proving any of those three conditions would disprove the null hypothesis.

The alternative hypothesis is β1≠0 OR β2≠0 OR β3≠0, via De Morgan's Laws. Proving any of those three conditions would disprove the null hypothesis.

The breakdown β1≠β2 OR β2≠β3 OR β3≠0 is mathematically equivalent but might be easier to show significance or design an experiment for.

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source | link

The alternative hypothesis is β1≠0 OR β2≠0 OR β3≠0, via De Morgan's Laws. Proving any of those three conditions would disprove the null hypothesis.