= Solution
Both relators have zero exponent sums, so <abelianization> gives $H_1(Y)=\mathbb Z^3$, with meridian variables $x,y,z$ corresponding to $a,b,c$. The <universal abelian cover> has <deck transformation group> $\mathbb Z^3$ and coefficient <group ring>
$$
R=\mathbb Z[x^{\pm1},y^{\pm1},z^{\pm1}].
$$
The <generalized Heegaard diagram> gives a two-dimensional spine with one vertex, three edges and two faces. Its lifted <cellular chain complex> is
$$
0\longrightarrow R^2\xrightarrow{d_2}R^3\xrightarrow{d_1}R\longrightarrow0,
$$
where chosen lifts of the cells give
$$
d_1=\begin{pmatrix}x-1&y-1&z-1\end{pmatrix},\qquad
d_2=\begin{pmatrix}
1-yz&1-z\\
x-1&x(1-z)\\
y(x-1)&xy-1
\end{pmatrix}.
$$
The two columns are the abelianized <Fox derivatives> of $[a,bc]$ and $[ab,c]$. The <Fox calculus> identity gives $d_1d_2=0$; this can also be checked by multiplying the displayed matrices. There is no three-cell in this spine. One may use the lifted spine because its <deformation retraction> from $Y$ lifts to the <universal abelian cover>.
The maximal minors of the <Alexander matrix>, in row-pair order $(1,2),(1,3),(2,3)$, are
$$
(z-1)(xyz-1),\quad -(y-1)(xyz-1),\quad (x-1)(xyz-1).
$$
Their <greatest common divisor> in $R$ is $xyz-1$, since $x-1,y-1,z-1$ have no common nonunit divisor. Accordingly the <multivariable Alexander polynomial> is
$$
\boxed{\Delta_L(x,y,z)\doteq xyz-1.}
$$
The allowed units are $\pm x^iy^jz^k$. The single-variable specialization convention can introduce extra $(t-1)$ factors; the answer here is the genuinely <multivariable Alexander polynomial>.
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