True or False: If the determinant $\small\begin{vmatrix}x+a & p+u & l+ f\\y+b & q+v & m +g\\z+c & r+w & n + h\end{vmatrix}$ splits into exactly $k$ determinants of order $3$, each element of which contains online term,then the value of $k$ is $8$.

$\begin{array}{1 1} True \\ False \end{array}$

Solution : True
Since, $\begin{vmatrix} x + a & p +u & l +f \\ y+b &q + v & m+g \\ z + c & r+w & n+h \end{vmatrix}$
$= \begin{vmatrix} x & p & l \\y+b & q+v & m+g \\ z+c & r+w & n+h \end{vmatrix} + \begin{vmatrix} a & u & f \\ y+b & q+v & m+g \\ z+c & r+w & n+h \end{vmatrix} \; \; \; \;$(splitting first row)
$=\begin{vmatrix}x & p & l \\ y & q&m \\ z+c & r+m & n+h \end{vmatrix} + \begin{vmatrix}x & p & l \\ b & v & g \\ z+c & r+w & n+h \end{vmatrix} + \begin{vmatrix} a & u & f \\ y & q & m \\ z+c & r+w & n+h \end{vmatrix} + \begin{vmatrix}a&u&f \\ b&v&g \\ z+c & r+w & n+h \end{vmatrix} (splitting\; second\; row)$
Similarly, we can split these 4 determinants in 8 determinants by splitting each one in two determinants further. SO, given statement is true.