= Solution
Put the three wires at $(x,y)=(-d,0),(0,0),(d,0)$ with $d=1\,\mathrm{cm}$ and each current along $+z$, out of the page. Adding their <Biot-Savart law> <magnetic fields> gives
$$
\mathbf B(x,y)=\frac{\mu_0I}{2\pi}\sum_{j=-1}^{1}
\frac{-y\hat{\mathbf x}+(x-jd)\hat{\mathbf y}}{(x-jd)^2+y^2}.
$$
For $y\ne0$, all three $x$-components have the same sign, so they cannot cancel. On $y=0$, a zero of the remaining component satisfies
$$
\frac1{x+d}+\frac1x+\frac1{x-d}
=\frac{3x^2-d^2}{x(x^2-d^2)}=0.
$$
Excluding the wire positions themselves, the two and only two finite zeros are
$$
\boxed{(x,y)=\left(\pm\frac d{\sqrt3},0\right)
=\left(\pm0.577\,\mathrm{cm},0\right).}
$$
The <magnetic field lines> circulate counterclockwise about each current near that wire. Far away they circle the combined total current $3I$. The two nulls separate the nearby circulation regions in the original <magnetic field line> plot below.
\Image[/past-exam-of-the-physics-2-paper-of-the-university-of-oxford/2017-three-wire-magnetic-field.png]
{title=Magnetic field and two nulls for three equal parallel currents}
{description=Coordinates are measured from the middle wire in units of the neighbouring-wire spacing. All three currents point out of the page; the two field zeros are marked between the wires.}
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