The value of $\left|\begin{array}{ccc}1 & 1 & 1 \\ { }^{n} C_{1} & { }^{n+2} C_{1} & { }^{n+4} C_{1} \\ { }^{n} C_{2} & { }^{n+2} C_{2} & { }^{n+4} C_{2}\end{array}\right|$ is
(a) 2
(b) 4
(c) 8
(d) $n^{2}$
$\left|\begin{array}{ccc}1 & 1 & 1 \\ { }^{n} C_{1} & { }^{n+2} C_{1} & { }^{n+4} C_{1} \\ { }^{n} C_{2} & { }^{n+2} C_{2} & { }^{n+4} C_{2}\end{array}\right|$
$=\left|\begin{array}{ccc}1 & 1 & 1 \\ n & n+2 & n+4 \\ \frac{n(n-1)}{2} & \frac{(n+2)(n+1)}{2} & \frac{(n+4)(n+3)}{2}\end{array}\right|$
$=\left|\begin{array}{ccc}1 & 0 & 0 \\ n & 2 & 4 \\ \frac{n(n-1)}{2} & \frac{4 n+2}{2} & \frac{8 n+12}{2}\end{array}\right|$ [Applying $C_{2} \rightarrow C_{2}-C_{1}$ and $C_{3} \rightarrow C_{3}-C_{1}$ ]
$=\left|\begin{array}{ccc}1 & 0 & 0 \\ n & 2 & 4 \\ \frac{n(n-1)}{2} & (2 n+1) & (4 n+6)\end{array}\right|$
$=8 n+12-8 n-4$
$=8$
Hence, the correct option is (c).
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