The rank of the matrix $\left[\begin{array}{ccc}-1& 2& 5\\ 2& -4& \mathrm{a}-4\\ 1& -2& \mathrm{a}+1\end{array}\right]\text{ is }1\text{ then }\left|a\right|=$

The given matrix is $\left[\begin{array}{ccc}-1& 2& 5\\ 2& -4& \mathrm{a}-4\\ 1& -2& \mathrm{a}+1\end{array}\right]$

Use the Elementary row operations to reduce the above matrix in Echlon form

${\mathrm{R}}_2\to {\mathrm{R}}_2+2{\mathrm{R}}_1$ and ${\mathrm{R}}_3\to {\mathrm{R}}_3+{\mathrm{R}}_1$

The equivalent matrix to the given matrix is 

$~\left[\begin{array}{ccr}-1& 2& 5\\ 0& 0& \mathrm{a}+6\\ 0& 0& \mathrm{a}+6\end{array}\right]$

Given that the rank of the matrix is one, Hence the number of non zero rows in the final matrix is only one

So that $a+6=0$

Therefore, $\left|a\right|=\boxed{\mathbf{6}}$