The value of the determinant $\left|\begin{array}{l}ka\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}{k}^2+a^2\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}1\\ kb\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}{k}^2+b^2\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}1\\ kc\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}{k}^2+c^2\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}1\end{array}\right|,$ is

We have,

$\mathrm{\Delta }=\left|\begin{array}{l}ka\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}{k}^2+a^2\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}1\\ kb\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}{k}^2+b^2\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}1\\ kc\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}{k}^2+c^2\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}1\end{array}\right|=\left|\begin{array}{l}ka\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}{k}^2\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}1\\ kb\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}{k}^2\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}1\\ kc\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}{k}^2\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}1\end{array}\right|+\left|\begin{array}{l}ka\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}{a}^2\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}1\\ kb\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}{b}^2\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}1\\ kc\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}{c}^2\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}1\end{array}\right|$

$\Rightarrow \mathrm{\Delta }=0+k\left|\begin{array}{l}a\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}{a}^2\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}1\\ b\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}{b}^2\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}1\\ c\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}{c}^2\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}1\end{array}\right|=k(a-b)(b-c)(c-a).$