Let $I$ be the identity matrix of order $3 \times 3$ and for the matrix $A=\left[\begin{array}{ccc}\lambda & 2 & 3 \\ 4 & 5 & 6 \\ 7 & -1 & 2\end{array}\right],|A|=-1$. Let $B$ be the inverse of the matrix $\operatorname{adj}\left(\operatorname{Aadj}\left(A^2\right)\right)$. Then $|(\lambda \mathrm{B}+\mathrm{I})|$ is equal to______

<p>$$\begin{aligned} & B=\left[\operatorname{adj}\left(A \operatorname{adj}\left(A^2\right)\right)\right]^{-1} \\ & \operatorname{Adj}\left(A^2\right)=(\operatorname{adj} A)^2 \Rightarrow A \operatorname{adj}\left(A^2\right)=A \operatorname{adj}(A) \cdot(\operatorname{adj} A) \\ & =A\left(|A| A^{-1}\right)^2=|A|^2\left(A^{-1}\right)=A^{-1} \\ & \Rightarrow B=\left(\operatorname{adj}\left(A^{-1}\right)\right)^{-1}=\left(\left|\left(A^{-1}\right)\right| A\right)^{-1}=\frac{A^{-1}}{-1}=-A^{-1} \\ & \Rightarrow B=-A^{-1} \end{aligned}$$</p> <p>$$\begin{aligned} & |A|=-1=\left|\begin{array}{lll} \lambda & 2 & 3 \\ 4 & 5 & 6 \\ 7 & -1 & 2 \end{array}\right|=-1 \Rightarrow \lambda=3 \\ & |3 B+I|=\left|I-3 A^{-1}\right|=\frac{|A|\left|I-3 A^{-1}\right|}{|A|}=\frac{|A-3 I|}{|A|} \\ & =\frac{|A-3 I|}{-1}=\frac{\left|\begin{array}{ccc} 0 & 2 & 3 \\ 4 & 2 & 6 \\ 7 & -1 & -1 \end{array}\right|}{-1}=38 \\ & \Rightarrow|3 B+I|=38 \end{aligned}$$</p>