Matrix of order \(3\times 3\) are formed by using the elements of the set \(A=\{-3,-2,-1,0,1,2,3\}\), then probability that matrix is either symmetric or skew symmetric is

\(\frac{1}{7^{6}}+\frac{1}{7^{3}}\) \(\frac{1}{7^{9}}+\frac{1}{7^{3}}-\frac{1}{7^{6}}\) \(\frac{1}{7^{3}}+\frac{1}{7^{9}}\) \(\frac{1}{7^{3}}+\frac{1}{7^{6}}-\frac{1}{7^{9}}\)

\(\frac{1}{7^{6}}+\frac{1}{7^{3}}\)

Favourable cases \(=7^{9}\)