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

Given,

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

$\overrightarrow{a}\times \overrightarrow{c}-\overrightarrow{a}\times \overrightarrow{b}=0$

$\Rightarrow \overrightarrow{a}\times \left(\overrightarrow{c}-\overrightarrow{b}\right)=0$

So, $\overrightarrow{a} \& \left(\overrightarrow{c}-\overrightarrow{b}\right)$ are parallel vectors,

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

$\Rightarrow \overrightarrow{c}=\overrightarrow{b}+\lambda \overrightarrow{a}$

$\Rightarrow \overrightarrow{a}·\overrightarrow{c}=\overrightarrow{a}·\overrightarrow{b}+\lambda {\left|\overrightarrow{a}\right|}^2$

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

$\Rightarrow 2\alpha +87\lambda =-29 ....\left(i\right)$

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

$\overrightarrow{c}=\hat{i}\left(\alpha +6\lambda \right)+\hat{j}(11+9\lambda )+\hat{k}(-2+12\lambda )$

Also given $\overrightarrow{c}·\left(\hat{i}-2\hat{j}+\hat{k}\right)=5$

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

$\Rightarrow \alpha =29$

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

$\Rightarrow \lambda =-1$

Hence, $\overrightarrow{c}=23\hat{i}+2\hat{j}-14\hat{k}$

So, the value of $\overrightarrow{c}·(\hat{i}+\hat{j}+\hat{k})=23+2-14=11$