Consider the group $G=\langle a, b, c; a^2=b^3=c^5=abc\rangle^{\ast}$. Clearly, $\langle abc\rangle$, which is the same subgroup as $\langle a^2\rangle$, as $\langle b^3\rangle$, and as $\langle c^5\rangle$, is a central subgroup of $G$. As we know from the earlier part of the problem, quotienting out this subgroup gives $A_5$, which has order $60$. Hence, in order to prove the result it is sufficient to prove that $(abc)^{2}=1$ and that $G\not\cong A_5$.
A proof of these facts can be found on the Groupprops website. Their proof shows that $G$ contains a subgroup which is a homomorphic image of the dicyclic group $\langle x, y, z; x^2=y^2=z^5=xyz\rangle$, and the subgroup is embedded such that $x_0y_0z_0=abc$. In a dicyclic group, $(x_0y_0z_0)^2=1$, hence $(abc)^2=1$ as required. The subgroup is given by the following embedding.
$$
\begin{align*}
x_0&\mapsto a^{-1}bab^{-1}a\\
y_0&\mapsto a\\
z_0&\mapsto cac^{-3}
\end{align*}
$$
I will leave you to prove that $G\not\cong A_5$. There are a number of ways of doing this. For example, you could prove that $abc\neq 1$ in $G$, or you could "realise" $G$ as a group of matrices (hint: $\operatorname{SL}_2(5)$) or permutations$^{\dagger}$. Read the wikipedia page to get yourself started.
$^{\ast}$The presentation given in this answer differs from the one you gave. However, proving that $(abc)^2=1$ (which is the purpose of this answer) proves that these presentations define isomorphic groups. Can you see why this is?
$^{\dagger}$Another possible, elegant, way would have been to have found a non-trivial homomorphic image of $G$ which was not $A_5$. However, in reality $G$ has only one proper, non-trivial subgroup so this won't work!
