Multiplying first, second, third rows by $a,b,c$ and then taking $a,b,c$ common from first, second, third columns, we get,
$D=\left|\begin{array}{lll}{a}^2+\lambda & a^2& a^2\\ b^2& b^2+\lambda & b^2\\ c^2& c^2& c^2+\lambda \end{array}\right|$

Now, $R_1\leftrightarrow R_1+R_2+R_3$
$=\left(a^2+b^2+c^2+\lambda \right)\left|\begin{array}{lll}1& 1& 1\\ b^2& b^2+\lambda & b^2\\ c^2& c^2& c^2+\lambda \end{array}\right|$

Now, $C_2\leftrightarrow C_2-C_1 , C_3\to C_3-C_1$
$=\left(a^2+b^2+c^2+\lambda \right)\left|\begin{array}{lll}1& 0& 0\\ b^2& \lambda & 0\\ c^2& 0& \lambda \end{array}\right|={\lambda }^2\left(a^2+b^2+c^2+\lambda \right)$
Hence, $D_{\mathrm{m}\mathrm{i}\mathrm{n}}=4$ when $a=b=c=\lambda =1$