If the determinant $$x+a$$ $$p+u$$ $$l+fy+b$$ $$q+v$$ $$m+gz+c$$ $$r+w$$ $$n+hsplits$$ into exactly k determinants of order $$3$$, each element of which contains only one term, then the value of k is

Question

If the determinant $$x+a$$    $$p+u$$   $$l+fy+b$$  $$q+v$$    $$m+gz+c$$   $$r+w$$   $$n+hsplits$$ into exactly k determinants of order $$3$$, each element of which contains only one term, then the value of k is

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since,$$x+a$$    $$p+u$$   $$l+fy+b$$  $$q+v$$    $$m+gz+c$$   $$r+w$$   $$n+h=$$   x        p          $$ly+b$$  $$q+v$$    $$m+gz+c$$   $$r+w$$   $$n+h$$ $$+$$  a        u           $$fy+b$$  $$q+v$$    $$m+gz+c$$   $$r+w$$   $$n+hsplitting$$ first $$row=$$   x        p          l   y       q         $$mz+c$$   $$r+w$$   $$n+h$$ $$+$$   x        p          l   b       v         $$gz+c$$   $$r+w$$   $$n+h+$$  a        u           f  y         q        $$mz+c$$   $$r+w$$   $$n+h+$$  a        u           f  b         v          $$gz+c$$   $$r+w$$   $$n+h$$                                                                                                                                                splitting second rowSimilarly we can split these $$4$$ determinants in $$8$$ determinants by splitting each one in two determinants further. So, $$k=8$$

If the determinant $|\begin{array}{l} x + a p + u l + f \\ y + b q + v m + g \\ z + c r + w n + h \end{array}|$ splits into exactly k determinants of order 3, each element of which contains only one term, then the value of k is

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If the determinant x+a p+u l+fy+b q+v m+gz+c r+w n+hsplits into exactly k determinants of order 3, each element of which contains only one term, then the value of k is

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If the determinant $|\begin{array}{l} x + a p + u l + f \\ y + b q + v m + g \\ z + c r + w n + h \end{array}|$ splits into exactly k determinants of order 3, each element of which contains only one term, then the value of k is