Let → a = ^ i + 2 ^ j + 3 ^ k and → b = ^ i + ^ j - ^ k . If → c is...

Given,

\overset{\to}{a} = \overset{^}{i} + 2\overset{^}{j} + 3\overset{^}{k}, \overset{\to}{b} = \overset{^}{i} + \overset{^}{j} - \overset{^}{k}, $\overset{\to}{a} \cdot \overset{\to}{c} = 11$, $\overset{\to}{b} \cdot \left(\overset{\to}{a} \times \overset{\to}{c}\right) = 27$ and $\overset{\to}{b} \cdot \overset{\to}{c} = - \sqrt{3}\left|\overset{\to}{b}\right|$

Now finding, \left(\overset{\to}{b} \times \overset{\to}{a}\right)we get,</p><p>\left(\overset{\to}{b} \times \overset{\to}{a}\right) = \left|\begin{cases}\overset{^}{i} & \overset{^}{j} & \overset{^}{k} \\ 1 & 1 & - 1 \\ 1 & 2 & 3\end{cases}\right| = 5\overset{^}{i} - 4\overset{^}{j} + \overset{^}{k}

Let \overset{\to}{c} = {c}_{1}\overset{^}{i} + {c}_{2}\overset{^}{j} + {c}_{3}\overset{^}{k}

Now solving, $\overset{\to}{a} \cdot \overset{\to}{c} = 11$ we get,

${c}_{1} + 2{c}_{2} + 3{c}_{3} = 11 . . . . . . . \left(1\right)$

Now solving, $\overset{\to}{b} \cdot \overset{\to}{c} = - \sqrt{3}\left|\overset{\to}{b}\right|$ we get,

${c}_{1} + {c}_{2} - {c}_{3} = - \sqrt{3}\sqrt{3}$

${c}_{1} + {c}_{2} - {c}_{3} = - 3 . . . . . . . . . . \left(2\right)$

Now solving \left(\overset{\to}{b} \times \overset{\to}{a}\right) \cdot \overset{\to}{c} = 27$we get,</p><p>$5{c}_{1} - 4{c}_{2} + {c}_{3} = 27 . . . . . . . . . . . \left(3\right)$</p><p>Now on solving equation \left(1\right), \left(2\right) \text{ and } \left(3\right)$ we get,

\overset{\to}{c} = 3\overset{^}{i} - 2\overset{^}{j} + 4\overset{^}{k}

Hence, {\left|\overset{\to}{a} \times \overset{\to}{c}\right|}^{2} = {\left|\begin{cases}\overset{^}{i} & \overset{^}{j} & \overset{^}{k} \\ 1 & 2 & 3 \\ 3 & - 2 & + 4\end{cases}\right|}^{2}= {\left|14\overset{^}{i} + 5\overset{^}{j} - 8\overset{^}{k}\right|}^{2}

$⇒ | \overset{\to}{a} \times \overset{\to}{c}{|}^{2} = {14}^{2} + {5}^{2} + {8}^{2} = 285$