Let → a = 6 ^ i + 9 ^ j + 12 ^ k , → b = α ^ i + 11 ^ j - 2 ^ k and...

Given,

\overset{\to}{a} = 6\overset{^}{i} + 9\overset{^}{j} + 12\overset{^}{k}, \overset{\to}{b} = \alpha\overset{^}{i} + 11\overset{^}{j} - 2\overset{^}{k} and $\overset{\to}{c}$ be vectors such that $\overset{\to}{a} \times \overset{\to}{c} = \overset{\to}{a} \times \overset{\to}{b}$,

$\overset{\to}{a} \times \overset{\to}{c} - \overset{\to}{a} \times \overset{\to}{b} = 0$

$⇒ \overset{\to}{a} \times \left(\overset{\to}{c} - \overset{\to}{b}\right) = 0$

So, $\overset{\to}{a} \text{ and } \left(\overset{\to}{c} - \overset{\to}{b}\right)$ are parallel vectors,

Hence, $\lambda\overset{\to}{a} = \overset{\to}{c} - \overset{\to}{b}$

$⇒ \overset{\to}{c} = \overset{\to}{b} + \lambda\overset{\to}{a}$

$⇒ \overset{\to}{a} \cdot \overset{\to}{c} = \overset{\to}{a} \cdot \overset{\to}{b} + \lambda{\left|\overset{\to}{a}\right|}^{2}$

$⇒ - 12 = \left(6\alpha + 75\right) + \lambda\left(261\right)$

$⇒ 2\alpha + 87\lambda = - 29 . . . . ( i )$

Now again using $\overset{\to}{c} = \overset{\to}{b} + \lambda\overset{\to}{a}$ we get,

\overset{\to}{c} = \overset{^}{i}\left(\alpha + 6\lambda\right) + \overset{^}{j} ( 11 + 9\lambda ) + \overset{^}{k} ( - 2 + 12\lambda )

Also given \overset{\to}{c} \cdot \left(\overset{^}{i} - 2\overset{^}{j} + \overset{^}{k}\right) = 5

$⇒ \left(\alpha + 6\lambda\right) - 2 ( 11 + 9\lambda ) + ( - 2 + 12\lambda ) = 5$

$⇒ \alpha = 29$

So, $2\alpha + 87\lambda = - 29$

$⇒ \lambda = - 1$

Hence, \overset{\to}{c} = 23\overset{^}{i} + 2\overset{^}{j} - 14\overset{^}{k}

So, the value of \overset{\to}{c} \cdot ( \overset{^}{i} + \overset{^}{j} + \overset{^}{k} ) = 23 + 2 - 14 = 11