Let S ={√ n: 1 ≤ n ≤ 50 and n is odd }. Let a ∈ S and A = [1, 0, a; -1, 1, 0; - a, 0, 1]. If a ∈ S Σ det ( adj A )=100 λ, then λ is equal to:

Let $\displaystyle \mathrm{S}=\{\sqrt{\mathrm{n}}: 1 \leqslant \mathrm{n} \leqslant 50$ and n is odd $\displaystyle \}$.Let $\displaystyle \mathrm{a} \in \mathrm{S}$ and $\displaystyle \mathrm{A}=\left[\begin{array}{rcc}1 & 0 & \mathrm{a} \\ -1 & 1 & 0 \\ -\mathrm{a} & 0 & 1\end{array}\right]$.If $\displaystyle \underset{\mathrm{a} \in \mathrm{S}}{\Sigma} \operatorname{det}(\operatorname{adj} \mathrm{A})=100 \lambda$, then $\displaystyle \lambda$ is equal to :