$|c| = 1$, આપણી પાસે $|c|^{2} = 1$ અથવા $c_1^{2} + c_2^{2} + c_3^{2} = 1 ……(i)$
ફરીથી જ્યારે $c \perp a$ અને $c \perp b,$
આપણી પાસે $c.a = 0$ $\Rightarrow\,\,a_1c_1 + a_2c_2 + a_3c_3 = 0 …….(ii)$
અને $c.b = 0$ $\Rightarrow \,\,b_1c_1 + b_2c_2 + b_3c_3 = 0 ……(iii)$
પણ જયારે $a$ અને $b$ વચ્ચેનો ખૂણો $\pi/6$ ત્યારે
$a. b = a_1b_1 + a_2b_2 + a_3b_3$
$|a| |b| cos \pi/6 = a_1b_1 + a_2b_2 + a_3b_3$
$3/4 (a_1^2 + a_2^2 + a_3^2) (b_1^2 + b_2^2 + b_3^2) = (a_1b_1 + a_2b_2 + a_3b_3)^2 …..(iv)$
હવે ${\left| {\,\begin{array}{*{20}{c}} {{{\text{a}}_{\text{1}}}}&{{{\text{a}}_{\text{2}}}}&{{a_3}} \\ {{{\text{b}}_{\text{1}}}}&{{{\text{b}}_{\text{2}}}}&{{b_3}} \\ {{{\text{c}}_{\text{1}}}}&{{{\text{c}}_{\text{2}}}}&{{c_3}} \end{array}\,} \right|^2}\,\, = \,\,\left| {\,\begin{array}{*{20}{c}} {{a_1}}&{{a_2}}&{{a_3}} \\ {{b_1}}&{{b_2}}&{{b_3}} \\ {{c_1}}&{{c_2}}&{{c_3}} \end{array}\,} \right|\,\,\left| {\,\begin{array}{*{20}{c}} {{a_1}}&{{a_2}}&{{a_3}} \\ {{b_1}}&{{b_2}}&{{b_3}} \\ {{c_1}}&{{c_2}}&{{c_3}} \end{array}\,} \right|$
$ = \,\,\left| {\begin{array}{*{20}{c}} {a_1^2\,\, + \;\,a_2^2\,\, + \,\,a_3^2}&{{a_1}{b_1}\,\, + \;\,{a_2}{b_2}\,\, + \;\,{a_3}{b_3}}&0 \\ {{b_1}{a_1}\,\, + \;\,{b_2}{a_2}\,\, + \;\,{b_3}{a_3}}&{b_1^2\,\, + \;\,b_2^2\,\, + \,\,b_3^2}&0 \\ 0&0&1 \end{array}\,} \right|\,\,$ {${\left( i \right),\,\,\left( {ii} \right)}$ અને ${\left( {iii} \right)}$ નો ઉપયોગ કરતાં}
$ = \,\,\frac{1}{4}\,\,\left( {a_1^2,\,\,a_2^2,\,\,a_3^2} \right)\,\,\left( {b_1^2\,\, + \;\,b_3^2\,\, + \;\,b_3^2} \right)$ {${\left( {iv} \right)}$ નો ઉપયોગ કરતાં } $ = \,\,\frac{{\left( {\sum {a_1^2} } \right)\,\,\left( {\sum {b_1^2} } \right)}}{4}\,\, = \,\,\frac{1}{4}\,\,|\overline a {|^2}\,\,|b{|^2}$
જ્યાં ${\sum {a_1^2} }$${ = \,\,a_1^2\,\, + \,\,a_2^2\,\, + \;\,a_3^2\,}$ અને ${\sum {b_1^2} }$${ = \,\,b_1^2\,\, + \,\,b_2^2\,\, + \;\,b_3^2}$