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(D) The equation of a plane passing through the intersection of two planes $\vec{r} \cdot \vec{n}_1 = d_1$ and $\vec{r} \cdot \vec{n}_2 = d_2$ is give

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(D) The equation of a plane passing through the intersection of two planes $\vec{r} \cdot \vec{n}_1 = d_1$ and $\vec{r} \cdot \vec{n}_2 = d_2$ is given by $\vec{r} \cdot (\vec{n}_1 + \lambda \vec{n}_2) = d_1 + \lambda d_2$.Here,$\vec{n}_1 = 3\hat{i} - \hat{j} + \hat{k}$,$d_1 = 1$,$\vec{n}_2 = \hat{i} + 4\hat{j} - 2\hat{k}$,and $d_2 = 2$.The equation is $\vec{r} \cdot ((3+\lambda)\hat{i} + (-1+4\lambda)\hat{j} + (1-2\lambda)\hat{k}) = 1 + 2\lambda$.Since the plane passes through the point $\vec{a} = \hat{i} + 2\hat{j} - \hat{k}$,we substitute $\vec{r} = \vec{a}$:$(\hat{i} + 2\hat{j} - \hat{k}) \cdot ((3+\lambda)\hat{i} + (-1+4\lambda)\hat{j} + (1-2\lambda)\hat{k}) = 1 + 2\lambda$.$(3+\lambda) + 2(-1+4\lambda) - (1-2\lambda) = 1 + 2\lambda$.$3 + \lambda - 2 + 8\lambda - 1 + 2\lambda = 1 + 2\lambda$.$11\lambda = 1 + 2\lambda \implies 9\lambda = 1 \implies \lambda = 1/9$.Substituting $\lambda = 1/9$ into the equation:$\vec{r} \cdot ((3+1/9)\hat{i} + (-1+4/9)\hat{j} + (1-2/9)\hat{k}) = 1 + 2/9$.$\vec{r} \cdot (\frac{28}{9}\hat{i} - \frac{5}{9}\hat{j} + \frac{7}{9}\hat{k}) = \frac{11}{9}$.$\vec{r} \cdot (28\hat{i} - 5\hat{j} + 7\hat{k}) = 11$.

Find the vector equation of the plane passing through the intersection of the planes $\vec{r} \cdot (3\hat{i} - \hat{j} + \hat{k}) = 1$ and $\vec{r} \cdot (\hat{i} + 4\hat{j} - 2\hat{k}) = 2$,and passing through the point $\hat{i} + 2\hat{j} - \hat{k}$.

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