If the determinant $\left|\begin{array}{l}x+ap+ul+f\\ y+bq+vm+g\\ z+cr+wn+h\end{array}\right|$splits into exactly k determinants of order 3, each element of which contains only one term, then the value of k is

Solution:

since,$\left|\begin{array}{l}x+ap+ul+f\\ y+bq+vm+g\\ z+cr+wn+h\end{array}\right|$

$=\left|\begin{array}{l}xpl\\ y+bq+vm+g\\ z+cr+wn+h\end{array}\right|\text{ +}\left|\begin{array}{l}auf\\ y+bq+vm+g\\ z+cr+wn+h\end{array}\right|$$\left[\text{splitting first row}\right]$

$=\left|\begin{array}{l}xpl\\ yqm\\ z+cr+wn+h\end{array}\right|+\left|\begin{array}{l}xpl\\ bvg\\ z+cr+wn+h\end{array}\right|+\left|\begin{array}{l}auf\\ yqm\\ z+cr+wn+h\end{array}\right|\text{+}\left|\begin{array}{l}auf\\ bvg\\ z+cr+wn+h\end{array}\right|$

                                                                                                                                                $\left[\text{splitting second row}\right]$

Similarly we can split these 4 determinants in 8 determinants by splitting each one in two determinants further. So, $\text{k}=\text{8}$