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(A) The given planes are $\vec{r} \cdot(\hat{i}+\hat{j}+\hat{k})-1=0$ and $\vec{r} \cdot(2 \hat{i}+3 \hat{j}-\hat{k})+4=0$.The equation of any plane p

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$y-3z+6=0$

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(A) The given planes are $\vec{r} \cdot(\hat{i}+\hat{j}+\hat{k})-1=0$ and $\vec{r} \cdot(2 \hat{i}+3 \hat{j}-\hat{k})+4=0$.The equation of any plane passing through the line of intersection of these planes is given by:$[\vec{r} \cdot(\hat{i}+\hat{j}+\hat{k})-1] + \lambda[\vec{r} \cdot(2 \hat{i}+3 \hat{j}-\hat{k})+4] = 0$Rearranging the terms,we get:$\vec{r} \cdot[(1+2\lambda)\hat{i} + (1+3\lambda)\hat{j} + (1-\lambda)\hat{k}] + (4\lambda-1) = 0$  ..........$(1)$The normal vector to this plane is $\vec{n} = (1+2\lambda)\hat{i} + (1+3\lambda)\hat{j} + (1-\lambda)\hat{k}$.Since the required plane is parallel to the $x$-axis,its normal vector must be perpendicular to the $x$-axis. The direction vector of the $x$-axis is $\hat{i} = (1, 0, 0)$.Thus,$\vec{n} \cdot \hat{i} = 0$:$(1+2\lambda)(1) + (1+3\lambda)(0) + (1-\lambda)(0) = 0$$1+2\lambda = 0 \Rightarrow \lambda = -\frac{1}{2}$.Substituting $\lambda = -\frac{1}{2}$ into equation $(1)$:$\vec{r} \cdot[(1-1)\hat{i} + (1-\frac{3}{2})\hat{j} + (1+\frac{1}{2})\hat{k}] + (4(-\frac{1}{2})-1) = 0$$\vec{r} \cdot[0\hat{i} - \frac{1}{2}\hat{j} + \frac{3}{2}\hat{k}] - 3 = 0$Multiplying by $-2$:$\vec{r} \cdot[\hat{j} - 3\hat{k}] + 6 = 0$.In Cartesian form,where $\vec{r} = x\hat{i} + y\hat{j} + z\hat{k}$,this becomes:$y - 3z + 6 = 0$.

Find the equation of the plane passing through the line of intersection of the planes $\vec{r} \cdot(\hat{i}+\hat{j}+\hat{k})=1$ and $\vec{r} \cdot(2 \hat{i}+3 \hat{j}-\hat{k})+4=0$ and parallel to the $x$-axis.

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