What would result from the following reactions:

ZrH + H2O ->

PrH + H2O ->

TiH + H2O ->

they are supposed to produce hydrogen gas and hydroxides

We need to seriously look at the formulas in light of the correct oxidation numbers.

Metallic hydrides easily react with water. As a hydride, hydrogen has an oxidation number of -1 and the common oxidation number of zirconium is +4, so ZrH4 is the most common hydride of zirconium.

ZrH4 + 4H2O --> Zr(OH)4(s) + 4H2(g)

The most common oxidation state of praseodymium is +3. It does not exhibit an oxidation state of +1.
PrH3 + 3H2O --> Pr(OH)3(s) + 3H2

Titanium exhibits oxidation number of +2 and +4, most commonly. The only hydride of titanium is TiH2.

TiH2 + 2H2O --> Ti(OH)2(s) + 2H2(g)

ZrH + H2O -> H2 + ZrOH
PrH + H2O -> H2 + PrOH
TiH + H2O -> H2 + TiOH