Let $\Delta  = \begin{array}{*{20}{c}}
{{a^2}}&{{b^2}}&{{c^2}}\\
{{{\left( {a + \lambda } \right)}^2}}&{{{\left( {b + \lambda } \right)}^2}}&{{{\left( {c + \lambda } \right)}^2}}\\
{{{\left( {a – \lambda } \right)}^2}}&{{{\left( {b – \lambda } \right)}^2}}&{{{\left( {c – \lambda } \right)}^2}}
\end{array}$

apply ${R_2} \to {R_2} – {R_3}$

$\Delta  = \begin{array}{*{20}{c}}
{{a^2}}&{{b^2}}&{{c^2}}\\
{{{\left( {a + \lambda } \right)}^2} – {{\left( {a – \lambda } \right)}^2}}&{{{\left( {b + \lambda } \right)}^2} – {{\left( {b – \lambda } \right)}^2}}&{{{\left( {c + \lambda } \right)}^2} – {{\left( {c – \lambda } \right)}^2}}\\
{{{\left( {a – \lambda } \right)}^2}}&{{{\left( {b – \lambda } \right)}^2}}&{{{\left( {c – \lambda } \right)}^2}}
\end{array}$

$ = \begin{array}{*{20}{c}}
{{a^2}}&{{b^2}}&{{c^2}}\\
{4a\lambda }&{4b\lambda }&{4c\lambda }\\
{{{\left( {a – \lambda } \right)}^2}}&{{{\left( {b – \lambda } \right)}^2}}&{{{\left( {c – \lambda } \right)}^2}}
\end{array}$

                    ($\because $ ${\left( {x + y} \right)^2} – {\left( {x – y} \right)^2} = 4xy$)

Taking out $4$ common form $\,{R_2}$

$ = 4\begin{array}{*{20}{c}}
{{a^2}}&{{b^2}}&{{c^2}}\\
{a\lambda }&{b\lambda }&{c\lambda }\\
{{a^2} + {\lambda ^2} – 2a\lambda }&{{b^2} + {\lambda ^2} – 2b\lambda }&{{c^2} + {\lambda ^2} – 2c\lambda }
\end{array}$

Apply ${R_3} \to \left[ {{R_3} – \left( {{R_1} – 2{R_2}} \right)} \right]$

$ = 4\begin{array}{*{20}{c}}
{{a^2}}&{{b^2}}&{{c^2}}\\
{a\lambda }&{b\lambda }&{c\lambda }\\
{{\lambda ^2}}&{{\lambda ^2}}&{{\lambda ^2}}
\end{array}$ 

Taking out $\lambda $ common from ${{R_2}}$ and  from ${{R_3}}$.

$ = 4\lambda \left( {{\lambda ^2}} \right)\begin{array}{*{20}{c}}
{{a^2}}&{{b^2}}&{{c^2}}\\
a&b&c\\
1&1&1
\end{array}$

$ = k\lambda \begin{array}{*{20}{c}}
{{a^2}}&{{b^2}}&{{c^2}}\\
a&b&c\\
1&1&1
\end{array}$

$ \Rightarrow k = 4{\lambda ^2}$