Explanation:
C Given, \(\mathrm{Z}=\) Set of integers,
and \(A=\left\{(a, b): a^2+3 b^2=28, a, b \in Z\right\}\)
\(\mathrm{B}=\{(\mathrm{a}, \mathrm{b}): \mathrm{a}\lt \mathrm{b}, \mathrm{a}, \mathrm{b} \in \mathrm{Z}\}\)
Then, in set \(A, a^2+3 b^2=28\) satisfies the following numbers are given below
\(=\left\{\begin{array}{l}
(-1,-3),(-1,3),(1,-3),(1,3),(-4,-2),(-4,2) \\
(4,-2),(4,2),(5,1),(-5,-1),(5,-1),(-5,1)
\end{array}\right\}\)
And in \(B, a\lt b\),
Then,
\(\{(1,3),(-1,3),(-4,2),(-4,-2),(-5,-1),(-5,1)\}\)
So, \(\mathrm{A} \cap \mathrm{B}=\{(1,3),(-1,3),(-4,2),(-4,-2),(-5,-1)\),\(\} .\)
\((-5,1)\}\)Hence, the number of element in \(\mathrm{A} \cap \mathrm{B}\) is 6 .