The vector equation of the plane passing through the intersection of the planes
$\overrightarrow{\mathrm{r}} \cdot(\hat{i}+\hat{j}+\hat{k})=1$ and $\overrightarrow{\mathrm{r}} \cdot(\hat{i}-2 \hat{j})=-2$, and the point $(1,0,2)$ is :
Correct Option: , 2
Plane passing through intersection of plane is
$\{\vec{r} \cdot(\hat{i}+\hat{j}+\hat{k})-1\}+\lambda\{\vec{r} \cdot(\hat{i}-2 \hat{j})+2\}=0$
Passes through $\hat{\mathrm{i}}+2 \hat{\mathrm{k}}$, we get
$(3-1)+\lambda(1+2)=0 \Rightarrow \lambda=-\frac{2}{3}$
Hence, equation of plane is $3\{\vec{r} \cdot(\hat{i}+\hat{j}+\hat{k})-1\}-2\{\vec{r} \cdot(\hat{i}-2 \hat{j})+2\}=0$
$\Rightarrow \overrightarrow{\mathrm{r}} \cdot(\hat{\mathrm{i}}+7 \hat{\mathrm{j}}+3 \hat{\mathrm{k}})=7$
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