Find the lengths of the medians of the triangle with vertices $A(0,0,6),B(0,4,0)$ and $(6,0,0).$

Let $AD,BE$ and $CF$ be the medians of the given $△ABC$.
Since $AD$ is the median, $D$ is the mid-point of $BC$.

$\therefore$ coordinates of point $D=$ $\left(\frac{0+6}{2},\frac{4+0}{2},\frac{0+0}{2}\right)=\left(3,2,0\right)$

$AD=$  $\sqrt{{\left(0-3\right)}^2+{\left(0-2\right)}^2+{\left(6-0\right)}^2}=\sqrt{9+4+36}=\sqrt{49}=7$

Since $BE$ is the median, $E$ is the mid-point of $AC$.

$\therefore$ coordinates of point $E=$ $\left(\frac{0+6}{2},\frac{0+0}{2},\frac{6+0}{2}\right)=\left(3,0,3\right)$

$BE=$ $\sqrt{{\left(3-0\right)}^2+{\left(0-4\right)}^2+{\left(3-0\right)}^2}=\sqrt{9+16+9}=\sqrt{34}$

Since $CF$ is the median, $F$ is the mid point of $AB$.

$\therefore$ coordinates of point $F=$ $\left(\frac{0+0}{2},\frac{0+4}{2},\frac{6+0}{2}\right)=\left(0,2,3\right)$

Length of $CF=$ $\sqrt{{\left(6-0\right)}^2+{\left(0-2\right)}^2+{\left(0-3\right)}^2}=\sqrt{36+4+9}=\sqrt{49}=7$

Thus the lengths of the medians of $△ABC$ are $7,\sqrt{34}$ and $7$ units.