Let 
$\mathrm{S}=\left\{(\mathrm{x},\mathrm{y})\in \mathrm{\mathbb{N}}\times \mathrm{\mathbb{N}}:9(\mathrm{x}-3{)}^2+16(\mathrm{y}-4{)}^2⩽144\right\}$ and $\mathrm{T}=\left\{(\mathrm{x},\mathrm{y})\in \mathrm{ℝ}\times \mathrm{ℝ}:(\mathrm{x}-7{)}^2+(\mathrm{y}-4{)}^2⩽36\right\}$.
The $\mathrm{n}(\mathrm{S}\cap \mathrm{T})$ is equal to ……….. .

$\begin{array}{l}\mathrm{S}:\frac{(\mathrm{x}-3{)}^2}{16}+\frac{(\mathrm{y}-4{)}^2}{9}⩽1;\mathrm{x},\mathrm{y}\in \{1,2,3,\dots \dots \}\\ \mathrm{T}:(\mathrm{x}-7{)}^2+(\mathrm{y}-4{)}^2⩽36\mathrm{x},\mathrm{y}\in \mathrm{R}\\ \text{ Let }\mathrm{x}-3=\mathrm{x}:\mathrm{y}-4=\mathrm{y}\\ \mathrm{S}:\frac{{\mathrm{x}}^2}{16}+\frac{{\mathrm{y}}^2}{9}⩽1;\mathrm{x}\in \{-2,-1,0,1,\dots \dots \}\\ \mathrm{T}:(\mathrm{x}-4{)}^2+{\mathrm{y}}^2⩽36;\mathrm{y}\in \{-3,-2,-1,0,\dots \dots \}\end{array}$

$\begin{array}{l}\mathrm{S}\cap \mathrm{T}=(-2,0),(-1,0),\dots \dots \dots (4,0)\to \left(7\right)\\ (-1,1),(0,1),\dots \dots \dots \dots (3,1)\to \left(5\right)\\ (-1,-1),(0,-1),\dots \dots \dots (3,-1)\to \left(5\right)\\ (-1,2),(0,2),(1,2),(2,2)\to \left(4\right)\\ (-1,-2),(0,-2),(1,-2),(2,-2)\to \left(4\right)\\ (0,3)(0,-3)\to \left(2\right)\end{array}$