$c =( a \times c )+ b$
$\Rightarrow | c |^{2}=| a + c |^{2}+| b |^{2}+2( a \times c ) \cdot b$
$\Rightarrow 2[ a b c ]=| a \times c |^{2}+ 1 -| c |^{2}$
Let angle between vectors a and c is $'\theta'$
So, $[a \, b \, c] =\frac{1}{2} [| c |^{2} \sin ^{2} \theta-| c |^{2}+1]$
$=\frac{1}{2}\left[ 1 -| c |^{2} \cos ^{2} \theta\right]$ ...(i)
Now, $c \cdot( a \cdot c )=( a \times c )^{2}+ b \cdot( a \times c )$
$\Rightarrow [a \, b \, c ]=| c |^{2} \sin ^{2} \theta$ ...(ii)
From Eq. (i) and (ii)
$\frac{1}{2}\left\{ 1 -| c |^{2} \cos ^{2} \theta\right\}=| c |^{2} \sin ^{2} \theta$
$\Rightarrow | c|^{2}=\frac{1}{\cos ^{2} \theta+2 \sin ^{2} \theta}$
$\because[ a b c ]=| c |^{2} \sin ^{2} \theta=\frac{\sin ^{2} \theta}{\cos ^{2} \theta+2 \sin ^{2} \theta}=\frac{\sin ^{2} \theta}{1+\sin ^{2} \theta}$
$=\frac{1}{\operatorname{cosec}^{2} \theta+1}$
For maximum value of $[ a b c ], \text{cosec}^{2} \theta=1$,
so $[a \, b \, c]_{\max }=\frac{1}{2}$