= Solution
\b[Yes: the displayed presentation already has <p-deficiency> exactly one for $p=2$.] The words $x,y,z,xy,yz,zx$ are not proper powers in the <free group>. For the length-two words this follows directly from their distinct consecutive letters in a <cyclically reduced word>. Thus each relator's $2$-root exponent is precisely the exponent of $2$ in its displayed power. The relator weights, in the given order, are
$$
\frac12,\quad1,\quad\frac14,\quad\frac1{16},\quad\frac1{16},\quad\frac18.
$$
Their sum is $2$, giving
$$
\boxed{\operatorname{def}_2=3-\left(\frac12+1+\frac14+\frac1{16}+\frac1{16}+\frac18\right)=1.}
$$
In particular the infinitude criterion proves that this group is infinite, although the question only asks for the existence of the presentation and prime.
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