Let x₀ be the point of Local maxima of $$f(x) = \overrightarrow a .\left( {\overrightarrow b \times \overrightarrow c } \right)$$, where
$\overrightarrow a = x\widehat i - 2\widehat j + 3\widehat k$, $\overrightarrow b = - 2\widehat i + x\widehat j - \widehat k$, $\overrightarrow c = 7\widehat i - 2\widehat j + x\widehat k$. Then the value of
$$\overrightarrow a .\overrightarrow b + \overrightarrow b .\overrightarrow c + \overrightarrow c .\overrightarrow a $$ at x = x₀ is :

$f(x) = \overrightarrow a \,.\,(\overrightarrow b \times \overrightarrow c )$<br><br>$= \left[ {\overrightarrow a \,\overrightarrow b \,\overrightarrow c } \right]$<br><br>$$ = \left| {\matrix{ x &amp; { - 2} &amp; 3 \cr { - 2} &amp; x &amp; { - 1} \cr 7 &amp; { - 2} &amp; x \cr } } \right|$$<br><br>$= x({x^2} + 2) + 2( - 2a + 7) + 3(4 - 7x)$<br><br>$\Rightarrow f(x) = {x^3} - 27x + 26$<br><br>$f'(x) = 3{x^2} - 27$<br><br>For maxima or minima $f'(x) = 0$<br><br>$\therefore$ $3{x^2} - 27 = 0$<br><br>$\Rightarrow x = \pm \,3$<br><br>Now, $f''(x) = 6x$<br><br>f''(x) at x = $-$3 is 6($-$3) = $-$18 &lt; 0<br><br>$\therefore$ At x = $-$3 f(x) is maximum.<br><br>So, x₀ = $-$3<br><br>$\therefore$ $\overrightarrow a = - 3\widehat i - 2\widehat j + 3\widehat k$<br><br>$\overrightarrow b = - 2\widehat i - 3\widehat j - \widehat k$<br><br>$\overrightarrow c = 7\widehat i - 2\widehat j - 3\widehat k$<br><br>Now, $$\overrightarrow a \,.\,\overrightarrow b \, + \,\overrightarrow b \,.\,\overrightarrow c \, + \,\overrightarrow c \,.\,\overrightarrow a $$<br><br>$= (6 + 6 - 3) + ( - 14 + 6 + 3) + ( - 21 + 4 - 9)$<br><br>$= 9 - 5 - 26$<br><br>$= - 22$