The line $( k + 1 ) ^ { 2 } x + k y - 2 k ^ { 2 } - 2 = 0$ passes through a fixed point. Distance of this point (-2, -1) is-

$\begin{aligned} & (k+1)^{2} x+k y-2 k^{2}-2=0 \\ \Rightarrow & \left(k^{2}+2 k+1\right) x+k y-2 k^{2}-2=0 \\ \Rightarrow & k^{2} x+2 k x+x+k y-2 k^{2}-2=0 \\ \Rightarrow & k^{2}(x-2)+2 k x+x+k y-2=0 \\ \Rightarrow & k^{2}(x-2)+k(2 x+y)+x-2=0 \\ \Rightarrow & (x-2)\left(k^{2}+1\right)+k(2 x+y)=0 \\ \Rightarrow & x-2+\frac{k}{k^{2}+1}(2 x+y)=0\end{aligned}$

$\begin{array}{c} \Rightarrow x-2=6 \\ \Rightarrow x=2 \end{array}$

$\begin{array}{c} \text { Again, } \frac{k}{k^{2}+1}(2 x+y)=0 \\ \therefore y=-9 \\ \therefore(x, y)=(2,-4) \end{array}$

$\therefore$ distance betweren $(2,-4){,}(-2,-1)$

$\begin{array}{l} =\sqrt{(2+2)^{2}+(-4+1)^{2}} \\ =\sqrt{11+9}=5 \end{array}$