If $(-3, 11), (6, 2)$ and $($$k$$, 4)$ are collinear points, then $k$ is equal to

If $3$ points $P\left( { x }_{ 1 },{ y }_{ 1 } \right) , Q\left( { x }_{ 2 },{ y }_{ 2 } \right)$ and $R\left( { x }_{ 3 },{ y }_{ 3 } \right)$ are collinear, then slope ${ m }_{ PQ }=$ slope ${ m }_{ QR } =$ slope ${ m }_{ PR }$.

The given points are $P\left( { x }_{ 1 },{ y }_{ 1 } \right) =\left( 3,11 \right) , Q\left( { x }_{ 2 },{ y }_{ 2 } \right) =\left( 6,2 \right)$ and $R\left( { x }_{ 3 },{ y }_{ 3 } \right) =\left( k,4 \right)$.

Now if $3$ points $P\left( { x }_{ 1 },{ y }_{ 1 } \right) , Q\left( { x }_{ 2 },{ y }_{ 2 } \right)$ and $R\left( { x }_{ 3 },{ y }_{ 3 } \right)$ are collinear, then

slope ${ m }_{ PQ }=$ slope ${ m }_{ QR } =$ slope ${ m }_{ PR }$

Slope ${ m }_{ PQ }=\dfrac { { y }_{ 2 }-{ y }_{ 1 } }{ { x }_{ 2 }-{ x }_{ 1 } } =\dfrac { 4-11 }{ k-3 }$, slope ${ m }_{ PR }=\dfrac { { y }_{ 3 }-{ y }_{ 1 } }{ { x }_{ 3 }-{ x }_{ 1 } } =\dfrac { 4-2 }{ k-6 }$

$\therefore \dfrac { 4-11 }{ k-3 } =\dfrac { 4-2 }{ k-6 }$

$\Rightarrow 2k+6=-7k+42$

$\Rightarrow k=4$