The points which are not collinear are:

$$ Let\quad us\quad take\quad three\quad points\quad A,\quad B\quad \& \quad C.\\ Three\quad points\quad A,\quad B\quad \& \quad C\quad will\quad be\quad collinear\\ If\quad AB+BC=AC.\\ Three\quad points\quad A,\quad B\quad \& \quad C\quad will\quad not\quad be\quad collinear\\ If\quad AB+BC=AC.\\ Let\quad us\quad investigate\quad thr\quad given\quad options.\\ Option\quad A\longrightarrow The\quad points\quad are\\ A(0,1)\quad B(8,3)\quad \& \quad C(6,7).\\ \therefore \quad AB=\sqrt { { \left( 0-8 \right)  }^{ 2 }+{ \left( 1-3 \right)  }^{ 2 } } units=\sqrt { 68 } =8.25units,\\ \quad \quad \quad BC=\sqrt { { \left( 8-6 \right)  }^{ 2 }+{ \left( 3-7 \right)  }^{ 2 } } units=\sqrt { 28 } =5.29units,\\ and\quad AC=\sqrt { { \left( 0-6 \right)  }^{ 2 }+{ \left( 1-7 \right)  }^{ 2 } } units=\sqrt { 72 } =8.49units.\\ \therefore \quad AB+BC=\left( 8.25+5.29 \right) units.=13.54units.\\ So\quad AB+BC\neq AC.\\ \therefore \quad A,\quad B\quad \& \quad C\quad are\quad not\quad collinear.\\ Option\quad B\longrightarrow The\quad points\quad are\\ A(4,3)\quad B(5,1)\quad \& \quad C(1,9).\\ \therefore \quad AB=\sqrt { { \left( 4-5 \right)  }^{ 2 }+{ \left( 3-1 \right)  }^{ 2 } } units=\sqrt { 5 } =2.24units,\\ \quad \quad \quad BC=\sqrt { { \left( 5-1 \right)  }^{ 2 }+{ \left( 1-9 \right)  }^{ 2 } } units=\sqrt { 80 } =8.94units,\\ and\quad AC=\sqrt { { \left( 4-1 \right)  }^{ 2 }+{ \left( 3-9 \right)  }^{ 2 } } units=\sqrt { 45 } =6.71units.\\ \therefore \quad AB+AC=\left( 2.24+6.71 \right) units=8.95units.\\ So\quad AB+AC=BC.\\ \therefore \quad A,\quad B\quad \& \quad C\quad are\quad \quad collinear.\\ Option\quad C\longrightarrow The\quad points\quad are\\ A(2,5)\quad B(-1,2)\quad \& \quad C(4,7).\\ \therefore \quad AB=\sqrt { { \left( 2+1 \right)  }^{ 2 }+{ \left( 5-2 \right)  }^{ 2 } } units=\sqrt { 18 } units=4.24units,\\ \quad \quad \quad BC=\sqrt { { \left( -1-4 \right)  }^{ 2 }+{ \left( 2-7 \right)  }^{ 2 } } units=\sqrt { 50 } units=7.07units,\\ and\quad AC=\sqrt { { \left( 2-4 \right)  }^{ 2 }+{ \left( 5-7 \right)  }^{ 2 } } units=\sqrt { 8 } units=2.83units.\\ \therefore \quad AB+AC=\left( 4.24+2.83 \right) units=7.07units.\\ So\quad AB+AC=BC.\\ \therefore \quad A,\quad B\quad \& \quad C\quad are\quad \quad collinear.\\ Option\quad C\longrightarrow The\quad points\quad are\\ A(-3,2)\quad B(1,-2)\quad \& \quad C(9,-10).\\ \therefore \quad AB=\sqrt { { \left( -3-1 \right)  }^{ 2 }+{ \left( 2+2 \right)  }^{ 2 } } units=\sqrt { 32 } units=5.66units,\\ \quad \quad \quad BC=\sqrt { { \left( 1-9 \right)  }^{ 2 }+{ \left( -2+10 \right)  }^{ 2 } } units=\sqrt { 128 } units=11.31units,\\ and\quad AC=\sqrt { { \left( -3-9 \right)  }^{ 2 }+{ \left( 2+10 \right)  }^{ 2 } } units=\sqrt { 288 } units=16.97units.\\ \therefore \quad AB+BC=\left( 5.66+11.31 \right) units=16.97units.\\ So\quad AB+BC=AC.\\ \therefore \quad A,\quad B\quad \& \quad C\quad are\quad \quad collinear.\\ Ans-\quad Option\quad A.\\ \\ \\  $$