If we consider a regular $n$-gon in plane, then if $s_1,s_2$ denote the reflection symmetries of this polygon with lines of reflections very closely chosen, then we obtain the presentation of the dihedral group as $$\langle s_1,s_2: s_1^2=s_2^2=(s_1s_2)^n=1\rangle.$$ Now consider tetrahedron. It can be shown that the full symmetry group of this is $S_4$, being permutation group of vertices.
But, do anybody have a simple idea to draw pictorially the three planes of reflections $s_1,s_2,s_2$ which yield the following relations: $$\langle s_1,s_2,s_2 : s_1^2=s_2^2=s_3^2=(s_1s_3)^2=(s_1s_2)^3=(s_2s_3)^3=1\rangle$$
I think a three dimensional representation of $S_4$ will do the trick. The elements $s_1, s_2$ and $s_3$ are represented by the matrices: $$ M_1 = \begin{pmatrix} 0 & -1 & 0 \\ -1 & 0 & 0 \\ 0 & 0 & 1 \end{pmatrix}, M_2 = \begin{pmatrix} 0 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 0 \end{pmatrix}, M_3 = \begin{pmatrix} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{pmatrix} $$ One can verify that they satisfy the same relationships as in the question. To find the visual components one has to look at the eigenvectors with eigenvalue $-1$, being the normals to the reflecting planes, they are $$ n_1= \begin{pmatrix} 1 \\ 1\\ 0 \end{pmatrix}, n_2= \begin{pmatrix} 1 \\ 0\\ -1 \end{pmatrix}, n_3= \begin{pmatrix} 1 \\ -1\\ 0 \end{pmatrix} $$