If $A=\left[\begin{array}{lll}x & 5 & 2 \\ 2 & y & 3 \\ 1 & 1 & z\end{array}\right], x y z=80,3 x+2 y+10 z=20$ and $A \operatorname{adj} A=k l$, then $k=$__________
Given:
$A=\left[\begin{array}{lll}x & 5 & 2 \\ 2 & y & 3 \\ 1 & 1 & z\end{array}\right]$
$x y z=80$
$3 x+2 y+10 z=20$
$A(\operatorname{adj} A)=k l$
Now,
$A=\left[\begin{array}{lll}x & 5 & 2 \\ 2 & y & 3 \\ 1 & 1 & z\end{array}\right]$
$\Rightarrow|A|=\left|\begin{array}{lll}x & 5 & 2 \\ 2 & y & 3 \\ 1 & 1 & z\end{array}\right|$
$\Rightarrow|A|=x(y z-3)-2(5 z-2)+1(15-2 y)$
$\Rightarrow|A|=x y z-3 x-10 z+4+15-2 y$
$\Rightarrow|A|=x y z-(3 x+2 y+10 z)+19$
$\Rightarrow|A|=80-(20)+19 \quad(\because x y z=80$ and $3 x+2 y+10 z=20)$
$\Rightarrow|A|=60+19$
$\Rightarrow|A|=79$
As we know,
$A(\operatorname{adj} A)=|A| I$
$\Rightarrow k I=79 I$
$\Rightarrow k=79$
Hence, $k=\underline{79}$.