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

$\vec{a} \cdot \vec{c}=-12, \vec{c} \cdot(\hat{i}-2 \hat{j}+\hat{k})=5$, then $\vec{c} \cdot(\hat{i}+\hat{j}+\hat{k})$ is equal to _______________.

Let $\vec{c}=c_1 \hat{i}+c_2 \hat{j}+c_3 \hat{k}$ <br/><br/>Now, $\vec{a} \cdot \vec{c}=-12$ <br/><br/>$\Rightarrow 6 c_1+9 c_2+12 c_3=-12$ ..............(i) <br/><br/>Also, $\vec{c} \cdot(\hat{i}-2 \hat{j}+\hat{k})=5$ <br/><br/>$\Rightarrow c_1-2 c_2+c_3=5$ ................(ii) <br/><br/>$$ \begin{aligned} & \text { Now, } \vec{a} \times \vec{c}=\vec{a} \times \vec{b} \\\\ & \Rightarrow \vec{a} \times(\vec{c}-\vec{b})=0 \\\\ & \Rightarrow \vec{a} \text { is parallel to }(\vec{c}-\vec{b}) \\\\ & \Rightarrow \vec{a}=\lambda(\vec{c}-\vec{b}) \\\\ & \Rightarrow 6 \hat{i}+9 \hat{j}+12 \hat{k}=\lambda\left(c_1-\alpha\right) \hat{i}+\lambda\left(c_2-11\right) \hat{j}+\lambda\left(c_3+2\right) \hat{k} \end{aligned} $$ <br/><br/>On comparing, we get <br/><br/>$c_1=\frac{6}{\lambda}+\alpha, c_2=\frac{9}{\lambda}+11, c_3=\frac{12}{\lambda}-2$ <br/><br/>Put there values in (ii), we get <br/><br/>$$ \begin{aligned} & \frac{6}{\lambda}+\alpha-\frac{18}{\lambda}-22+\frac{12}{\lambda}-2=5 \\\\ & \Rightarrow \alpha=29 \end{aligned} $$ <br/><br/>From (i) and values of $\mathrm{c}_1, \mathrm{c}_2, \mathrm{c}_3$, and $\alpha$ we have <br/><br/>$$ \begin{aligned} & 6\left(\frac{6}{\lambda}+29\right)+9\left(\frac{9}{\lambda}+11\right)+12\left(\frac{12}{\lambda}-2\right)=-12 \\\\ & \Rightarrow \frac{261}{\lambda}=-261 \Rightarrow \lambda=-1 \end{aligned} $$ <br/><br/>$$ \begin{aligned} & \text { So, } c_1=23, c_2=2, c_3=-14 \\\\ & \therefore \vec{c} \cdot(\hat{i}+\hat{j}+\hat{k})=(23 \hat{i}+2 \hat{j}+-14 \hat{k}) \cdot(\hat{i}+\hat{j}+\hat{k}) \\\\ & =23+2-14=11 \end{aligned} $$