Integral Co-ordinates -3

Lets take yhe vertices as $P(\alpha _1 , \alpha _2) , Q(\beta _1, \beta _2)$ and $R(\gamma _1 , \gamma _2)$

Let $L_1$ be the line joining $P$ and $Q$ , $L_2$ be line joining $Q$ and $R$ and $L_3$ be line joining $P$ and $R$

Hence, $L_1 \rightarrow \dfrac {x - \alpha _1}{\beta _1 - \alpha _1} = \dfrac {y - \alpha _2}{\beta _2 - \alpha _2}$

$L_2 \rightarrow \dfrac {x - \beta _1}{\gamma _1 - \beta _1} = \dfrac {y - \beta _2}{\gamma _2 - \beta _2}$

And $L_3 \rightarrow \dfrac {x - \gamma _1}{\gamma _1 - \alpha _1} = \dfrac {y - \gamma _2}{\gamma _2 - \alpha _2}$

Using Pick's Theorem , Area of Polygon, $A = I + \dfrac B2 -1$

Where, $I$ is number of points on the interior of the polygon

$B$ is the number of points on the boundary of the polygon.

$B=$ Number of points on the lines ${\color{#20A900}{L_1}},{\color{#3D99F6}{L_2}}$ and ${\color{#D61F06}{L_3}}$ including end points

$= {\color{#20A900}{gcd(\beta _2 - \alpha _2 , \beta _1 - \alpha _1) -1}} + {\color{#3D99F6}{gcd(\gamma _2 - \beta _2 , \gamma _1 - \beta _1) - 1}} + {\color{#D61F06}{gcd(\gamma _2 - \alpha _2 , \gamma _1 - \alpha _1) - 1}} + 3_{P, Q, R}$

Area of Triangle is given by $A'= \dfrac 12 \begin{vmatrix} \alpha _1 & \alpha _2 & 1 \\ \beta _1 & \beta _2 & 1 \\ \gamma _1 & \gamma _2 & 1 \end{vmatrix}$

Now, $A = A'$

$\boxed{{\color{#20A900}{N = I = \dfrac 12 \begin{vmatrix} \alpha _1 & \alpha _2 & 1 \\ \beta _1 & \beta _2 & 1 \\ \gamma _1 & \gamma _2 & 1 \end{vmatrix} - \dfrac 12 ( gcd(\beta _2 - \alpha _2 , \beta _1 - \alpha _1) + gcd(\gamma _2 - \beta _2 , \gamma _1 - \beta _1) + gcd(\gamma _2 - \alpha _2 , \gamma _1 - \alpha _1) ) + 1}}}$

Now, for our Coordinates input in the above formula to get $N = \boxed{3}$

Shifting the triangle in the co-ordinate plane would not change the number of lattice points since two similarly-oriented congruent triangles should have the same no. of lattice points. We can shift the triangle by subtracting equal x and y-coordinates from each vertex preserving the similar orientation and congruency. . So, required no. of points will be same as that inside $P'=(3,0);Q'=(0,3);R'=(2,4)$ -- x co.ordniate $-$ 2016,y co.ordinate $-$ 2017. Now either employ pick's theorem or if not use manual counting.