Simultaneous Equations in 4 variables!

Note that

$(a+b+c)+(a+b+d)+(a+c+d)+(b+c+d)$

$=3a+3b+3c+3d$

$=3(a+b+c+d)$

$=5+7+11+13$

$=36.$

Hence $a+b+c+d=\frac{36}{3}=12 \implies a=-1,b=1,c=5,d=7$ from the given equations. We have $\displaystyle|abcd\displaystyle|=\displaystyle|-35\displaystyle|=\boxed{35}$ and we are done.

Adding up all the 4 equations, we get:

$3 (a + b + c + d) = 36$ $a + b + c + d = 12$

With this equation, we subtract equations one at a time to get:

$a = -1, b=1, c=5, d =7$

Computing the answer:

$\begin{aligned} |(1)(-1)(5)(7)| &= |-35| \\ &= \boxed{35} \end{aligned}$