The vector equation of the plane passing through the intersection of the planes $\vec{r}.\left ( \hat{i}+\hat{j}+\hat{k} \right )=1$ and $\vec{r}.\left ( \hat{i}-2\hat{j} \right )=-2,$ and the point $\left ( 1,0,2 \right )$ is :   Option: 1 $\vec{r}.\left ( \hat{i}+7\hat{j}+3\hat{k} \right )=\frac{7}{3}$ Option: 2 $\vec{r}.\left ( \hat{i}-7\hat{j}+3\hat{k} \right )=\frac{7}{3}$ Option: 3 $\vec{r}.\left ( \hat{i}+7\hat{j}+3\hat{k} \right )=7$ Option: 4 $\vec{r}.\left ( 3\hat{i}+7\hat{j}+3\hat{k} \right )=7$

$\\\overrightarrow{\mathrm{r}} \cdot(\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}})=1 \\ \\\overrightarrow{\mathrm{r}} \cdot(\hat{\mathrm{i}}-2 \hat{\mathrm{j}})=-2 \\ \\\text { point }(1,0,2)$

$\mathrm{Eq}^{\mathrm{n}} \text { of plane }$

$\\\overrightarrow{\mathrm{r}} \cdot(\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}})-1+\lambda\{\mathrm{r} \cdot(\hat{\mathrm{i}}-2 \hat{\mathrm{j}})+2\}=0 \\ \\\overrightarrow{\mathrm{r}}\{\hat{\mathrm{i}}(1+\lambda)+\hat{\mathrm{j}}(1-2 \lambda)+\hat{\mathrm{k}}(1)\}-1+2 \lambda=0 \\ \\\text { Point } \hat{\mathrm{i}}+0 \hat{\mathrm{j}}+2 \hat{\mathrm{k}}=\overrightarrow{\mathrm{r}}$

$\\\therefore(\hat{\mathrm{i}}+2 \hat{\mathrm{k}}) \cdot\{\hat{\mathrm{i}}(1+\lambda)+\hat{\mathrm{j}}(1-2 \lambda)+\hat{\mathrm{k}}(1)\}-1+2 \lambda=0 \\ \\1+\lambda+2-1+2 \lambda=0 \\ \\\lambda=-\frac{2}{3}$

$\\\therefore \quad \overrightarrow{\mathrm{r}} \cdot\left[\hat{\mathrm{i}}\left(\frac{1}{3}\right)+\hat{\mathrm{j}}\left(\frac{7}{3}\right)+\hat{\mathrm{k}}\right]=\frac{7}{3}\\ \\\mathrm{r} \cdot[\hat{\mathrm{i}}+7 \hat{\mathrm{j}}+3 \hat{\mathrm{k}}]=7$

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