In the reduction done here, you take one clause $d = a \cup b\cup c$, and make 10 clauses
$a, b, c, d, a \cup \neg d, b \cup \neg d, c \cup \neg d, \neg a \cup \neg b, \neg b \cup \neg c, \neg a \cup \neg c$
Then in the case of all $a,b,c = 0$ leads to only 6 clauses being satisfied, while if any are true ($=1$), then 7 clauses are satisfied.
I see that all $0$ case leads to 6 satisfied, and any $2$ or all $3$ being True leads to 7 clauses satisfied, but when only one of $a,b,c$ are True, it looks like only 6 are satisfied.
Let $a=1, b=c=0$. Then in order of clauses above,
$1,0,0,1,1,0,0,1,1,1$ are the results of the clauses. This leads to 6 being satisfied. What did I do wrong?
$d$ is freely chosen. Letting $d=0$ then leads to 7 being satisfied.