Official solution
\[\begin{aligned}
& \lim _{x \rightarrow 0} f(x)=\left|\[\begin{array}{|ccccccc|}a+1 & 1 & b \\ a & 1+1 & b \\ a & 1 & b+1\end{array}\end{aligned}\]\right| \\
\& =(a+1)(2(b+1)-b)+1(a b-a(b+1))+b a \\
\& =(a+1)(b+2)-a+a b \\
\& =b+a+2=\lambda+\mu a+v b \\
\& \lambda=2, \mu=1, v=1 \Rightarrow(\lambda+\mu+v)\^2=16
\]