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(A) The equations of the planes are $\vec{r} \cdot (2\hat{i} + 2\hat{j} - 3\hat{k}) = 7$ and $\vec{r} \cdot (2\hat{i} + 5\hat{j} + 3\hat{k}) = 9$.The

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$\vec{r} \cdot (38\hat{i} + 68\hat{j} + 3\hat{k}) = 153$

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(A) The equations of the planes are $\vec{r} \cdot (2\hat{i} + 2\hat{j} - 3\hat{k}) = 7$ and $\vec{r} \cdot (2\hat{i} + 5\hat{j} + 3\hat{k}) = 9$.The equation of any plane passing through the intersection of these two planes is given by:$[\vec{r} \cdot (2\hat{i} + 2\hat{j} - 3\hat{k}) - 7] + \lambda [\vec{r} \cdot (2\hat{i} + 5\hat{j} + 3\hat{k}) - 9] = 0$,where $\lambda \in \mathbb{R}$.Rearranging the terms,we get:$\vec{r} \cdot [(2 + 2\lambda)\hat{i} + (2 + 5\lambda)\hat{j} + (-3 + 3\lambda)\hat{k}] = 7 + 9\lambda$  ... $(1)$Since the plane passes through the point $(2,1,3)$,its position vector is $\vec{r} = 2\hat{i} + \hat{j} + 3\hat{k}$.Substituting this into equation $(1)$:$(2\hat{i} + \hat{j} + 3\hat{k}) \cdot [(2 + 2\lambda)\hat{i} + (2 + 5\lambda)\hat{j} + (3\lambda - 3)\hat{k}] = 7 + 9\lambda$Calculating the dot product:$2(2 + 2\lambda) + 1(2 + 5\lambda) + 3(3\lambda - 3) = 7 + 9\lambda$$4 + 4\lambda + 2 + 5\lambda + 9\lambda - 9 = 7 + 9\lambda$$18\lambda - 3 = 7 + 9\lambda$$9\lambda = 10 \Rightarrow \lambda = \frac{10}{9}$.Substituting $\lambda = \frac{10}{9}$ into equation $(1)$:$\vec{r} \cdot [(2 + 2(\frac{10}{9}))\hat{i} + (2 + 5(\frac{10}{9}))\hat{j} + (3(\frac{10}{9}) - 3)\hat{k}] = 7 + 9(\frac{10}{9})$$\vec{r} \cdot [(\frac{18+20}{9})\hat{i} + (\frac{18+50}{9})\hat{j} + (\frac{30-27}{9})\hat{k}] = 7 + 10$$\vec{r} \cdot [\frac{38}{9}\hat{i} + \frac{68}{9}\hat{j} + \frac{3}{9}\hat{k}] = 17$Multiplying by $9$,we get:$\vec{r} \cdot (38\hat{i} + 68\hat{j} + 3\hat{k}) = 153$.

Find the vector equation of the plane passing through the intersection of the planes $\vec{r} \cdot(2 \hat{i}+2 \hat{j}-3 \hat{k})=7$ and $\vec{r} \cdot(2 \hat{i}+5 \hat{j}+3 \hat{k})=9$ and passing through the point $(2,1,3).$

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