On the off chance that we need to demonstrate that the given three focuses \[\left( \mathbf{3},\text{ }\mathbf{0} \right),\text{ }\left( \text{ }\mathbf{2},\text{ }\text{ }\mathbf{2} \right)\text{ }\mathbf{and}\text{ }\left( \mathbf{8},\text{ }\mathbf{2} \right)\]  are collinear, then, at that point, we need to likewise demonstrate that the line going through the focuses \[\left( \mathbf{3},\text{ }\mathbf{0} \right)\text{ }\mathbf{and}\text{ }\left( \text{ }\mathbf{2},\text{ }\text{ }\mathbf{2} \right)\] additionally goes through the point \[\left( \mathbf{8},\text{ }\mathbf{2} \right).\]

By utilizing the equation,

The condition of the line going through the focuses \[\left( \mathbf{x1},\text{ }\mathbf{y1} \right)\text{ }\mathbf{and}\text{ }\left( \mathbf{x2},\text{ }\mathbf{y2} \right)\] is given by

\[-\text{ }\mathbf{5y}\text{ }=\text{ }-\text{ }\mathbf{2}\text{ }\left( \mathbf{x}\text{ }-\text{ }\mathbf{3} \right)\]

\[-\text{ }\mathbf{5y}\text{ }=\text{ }-\text{ }\mathbf{2x}\text{ }+\text{ }\mathbf{6}\]

\[\mathbf{2x}\text{ }\text{ }\mathbf{-5y}\text{ }=\text{ }\mathbf{6}\]

On the off chance that \[\mathbf{2x}\text{ }\text{ }\mathbf{-5y}\text{ }=\text{ }\mathbf{6}\]  goes through \[\left( \mathbf{8},\text{ }\mathbf{2} \right),\]

\[\mathbf{2x}\text{ }\text{ }\mathbf{-5y}\text{ }=\text{ }\mathbf{2}\left( \mathbf{8} \right)\text{ }\text{ }\mathbf{-5}\left( \mathbf{2} \right)\]

\[=\text{ }\mathbf{16}\text{ }-\text{ }\mathbf{10}\]

\[=\text{ }\mathbf{6}\]

\[=\text{ }\mathbf{RHS}\]

The line going through the focuses \[\left( \mathbf{3},\text{ }\mathbf{0} \right)\text{ }\mathbf{and}\text{ }\left( \text{ }\mathbf{2},\text{ }\text{ }\mathbf{2} \right)\]  likewise goes through the point \[\left( \mathbf{8},\text{ }\mathbf{2} \right).\]

Thus demonstrated. The given three focuses are collinear.