"Unpredictable Solution" Problem (Part 1)

$\begin{cases} abc + ab + ac + bc + a+ b + c = 7 \space ...(1) \\ bcd + bc + bd + cd + b + c + d = 9 \space ...(2) \\ cda + ca + cd + da + c + d + a = 9 \space ...(3) \\ cda + ca + cd + da + c + d + a = 9\space ...(4) \end{cases}$

$(1) + 1 \Rightarrow \space abc + ab + ac + bc + a+ b + c + 1 = 8 \rightarrow (a+1)(b+1)(c+1) = 8 \space ...(5)$ .

$(2) + 1 \Rightarrow \space bcd + bc + bd + cd + b + c + d + 1= 10 \rightarrow (b+1)(c+1)(d+1) = 10 \space ...(6)$ .

$(3) + 1 \Rightarrow \space cda + ca + cd + da + c + d + a +1 = 10 \rightarrow (c+1)(d+1)(a+1) = 10 \space ...(7)$ .

$(4) + 1 \Rightarrow \space cda + ca + cd + da + c + d + a +1 = 10 \rightarrow (d+1)(a+1)(b+1) = 10 \space ...(8)$ .

$(5) \times (6) \times (7) \times (8) = \Rightarrow \space (a+1)³(b+1)³(c+1)³(d+1)³ = 8000 \rightarrow \space (a+1)(b+1)(c+1)(d+1) = 20 \space ...(9)$ .

$(9) : (1)$ gives $d = \frac{3}{2}$ .

$(9) : (2)$ gives $a =1$ .

$(9) : (3)$ gives $b =1$ .

$(9) : (4)$ gives $c = 1$ .

Then, we have $11a^2 + 10b^2 + 9c^2 + 8d^2 + 7ab + 6bc + 5cd + 4da + 3db + 2ac + abcd + \frac{1}{2}$

$= 11 + 10 + 9 + 18 + 7 + 6 + \frac{15}{2} + 6 + \frac{9}{2} + 2 + \frac{3}{2} + \frac{1}{2}$

Hence, our final answer $\boxed{83}$ .