The greatest value of n for which the determinant

$Δ=\begin{vmatrix} 1 & 1 & 1 \\ {^nC_1} & {^{n+3}C_1}& {^{n+6}C_1}\\ {^nC_2} & {^{n+3}C_2}& {^{n+6}C_2}\end{vmatrix} $ is divisible by $3^n$ , is

3

The correct answer is option (4) : 3

We have,

$Δ =\frac{1}{2}\begin{vmatrix} 1 & 1& 1\\n & n+3 & n+6 \\ n(n-1) & (n+2)(n+3) & (n+5)(n+6)\end{vmatrix}$

$Δ =\frac{1}{2}\begin{vmatrix} 1 & 0 & 0\\n & 3 & 6 \\ n(n-1) & 6n+6 & 12n+30\end{vmatrix}$

$Δ =9\begin{vmatrix} 1 & 0 & 0\\n & 1 & 1 \\ n(n-1) & 2(n+1) & 2n+5\end{vmatrix}=27$

Clearly, $Δ= 27 $ is divisible by $3, 3^2 $ and $3^3$. Hence, $n=3 .$