Given ,
${\mathrm{ax}}_1^2+{\mathrm{by}}_1^2+{\mathrm{cz}}_1^2={\mathrm{ax}}_2^2+{\mathrm{by}}_2^2+{\mathrm{cz}}_2^2={\mathrm{ax}}_3^2+{\mathrm{by}}_3^2+{\mathrm{cz}}_3^2=\mathrm{d}$                  .....(1)

ax₂x₃ + by₂y₃ + cz₂z₃ = ax₃x₁ + by₃y₁ + cz₃z₁ + ax₁x₂ + by₁y₂ + cz₁z₂ = f          ....(2)
Let $∆=\mid \begin{array}{ccc}{\mathrm{x}}_1& {\mathrm{y}}_1& {\mathrm{z}}_1\\ {\mathrm{x}}_2& {\mathrm{y}}_2& {\mathrm{z}}_2\\ {\mathrm{x}}_3& {\mathrm{y}}_3& {\mathrm{z}}_3\end{array}\mid$
${∆}^2=\frac{1}{\mathrm{abc}}\mid \begin{array}{ccc}{\mathrm{x}}_1& {\mathrm{y}}_1& {\mathrm{z}}_1\\ {\mathrm{x}}_2& {\mathrm{y}}_2& {\mathrm{z}}_2\\ {\mathrm{x}}_3& {\mathrm{y}}_3& {\mathrm{z}}_3\end{array}\mid \times \mid \begin{array}{ccc}{\mathrm{ax}}_1& {\mathrm{by}}_1& {\mathrm{cz}}_1\\ {\mathrm{ax}}_2& {\mathrm{by}}_2& {\mathrm{cz}}_2\\ {\mathrm{ax}}_3& {\mathrm{by}}_3& {\mathrm{cz}}_3\end{array}\mid$
$=\frac{1}{\mathrm{abc}}\mid \begin{array}{ccc}\mathrm{d}& \mathrm{f}& \mathrm{f}\\ \mathrm{f}& \mathrm{d}& \mathrm{f}\\ \mathrm{f}& \mathrm{f}& \mathrm{d}\end{array}\mid$   (by (1) and (2))
C₁ → C₁ + C₂ + C₃
$=\frac{1}{\mathrm{abc}}\mid \begin{array}{ccc}\mathrm{d}+2\mathrm{f}& \mathrm{f}& \mathrm{f}\\ \mathrm{d}+2\mathrm{f}& \mathrm{d}& \mathrm{f}\\ \mathrm{d}+2\mathrm{f}& \mathrm{f}& \mathrm{d}\end{array}\mid =\frac{(\mathrm{d}+2\mathrm{f})}{\mathrm{abc}}\mid \begin{array}{ccc}1& \mathrm{f}& \mathrm{f}\\ 1& \mathrm{d}& \mathrm{f}\\ 1& \mathrm{f}& \mathrm{d}\end{array}\mid$

On solving it we get
${\mathrm{\Delta }}^2=\frac{(\mathrm{d}+2\mathrm{f})}{\mathrm{abc}}(\mathrm{d}-\mathrm{f}{)}^2$
$\mathrm{\Delta }=(\mathrm{d}-\mathrm{f}){\left[\frac{\mathrm{d}+2\mathrm{f}}{\mathrm{abc}}\right]}^{\frac{1}{2}}$