= Solution
Take the <kaon> decay as the common origin. A <pion> traveling freely with constant momentum has lab decay time $t_i=(E_i/m_\pi c^2)\tau_i$ and position $\mathbf r_i=(\mathbf p_i/m_\pi)\tau_i$. The supplied proper survival times therefore give
$$
\boxed{t_1=22.1710\ \mathrm{ns},\quad t_2=84.5978\ \mathrm{ns},\quad
\mathbf r_1=(0,4.89455)\ \mathrm m,\quad
\mathbf r_2=(21.59875,-9.78910)\ \mathrm m.}
$$
Consequently the separation from the first decay event to the second is
$$
\boxed{\Delta t=62.4268\ \mathrm{ns},\quad
\Delta\mathbf r=(21.59875,-14.68365)\ \mathrm m,\quad
|\Delta\mathbf r|=26.1173\ \mathrm m.}
$$
The requested distance is the separation of the two events, not the sum of the two individual <pion> path lengths.
Here $|\Delta\mathbf r|>c|\Delta t|$, so the separation is a <spacelike separation>. There is an <inertial frame> in which the events are simultaneous; in that frame their distance equals the invariant spacelike length:
$$
\boxed{d_{\rm simultaneous}=\sqrt{|\Delta\mathbf r|^2-c^2\Delta t^2}
=18.2170\ \mathrm m.}
$$
One such frame moves along $\Delta\mathbf r$ with $\mathbf V=c^2\Delta t\,\Delta\mathbf r/|\Delta\mathbf r|^2$, of speed $0.71658c$. The <simultaneity> condition is then $\Delta t'=\gamma_V(\Delta t-\mathbf V\cdot\Delta\mathbf r/c^2)=0$.
The <spacelike distance between daughter decay events> provides an independent invariant check: the daughter <four-velocity> contraction gives
$$
\frac{d_{\rm simultaneous}^2}{c^2}
=\frac{M^2}{m_\pi^2}\tau_1\tau_2-(\tau_1+\tau_2)^2.
$$
It produces the same distance without using the particular <kaon> laboratory boost. A simultaneous frame exists only when this expression is positive, as it is for these two survival times.
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