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(D) The plane passes through $(1, 2, -3)$ and $(2, -2, 1)$. The vector connecting these two points is $\vec{v} = (2-1)\hat{i} + (-2-2)\hat{j} + (1-(-3

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$y + z + 1 = 0$

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(D) The plane passes through $(1, 2, -3)$ and $(2, -2, 1)$. The vector connecting these two points is $\vec{v} = (2-1)\hat{i} + (-2-2)\hat{j} + (1-(-3))\hat{k} = \hat{i} - 4\hat{j} + 4\hat{k}$.Since the plane is parallel to the $X$-axis, its normal is perpendicular to the unit vector $\hat{i} = (1, 0, 0)$.The normal vector $\vec{n}$ to the plane is given by the cross product of $\vec{v}$ and $\hat{i}$:$\vec{n} = \vec{v} 	imes \hat{i} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & -4 & 4 \ 1 & 0 & 0 \end{vmatrix} = \hat{i}(0) - \hat{j}(0-4) + \hat{k}(0-(-4)) = 4\hat{j} + 4\hat{k}$.We can simplify the normal vector to $\vec{n}' = (0, 1, 1)$.The equation of the plane passing through $(1, 2, -3)$ with normal $(0, 1, 1)$ is:$0(x-1) + 1(y-2) + 1(z+3) = 0$$y - 2 + z + 3 = 0$$y + z + 1 = 0$.

The equation of the plane passing through the points $(1, 2, -3)$ and $(2, -2, 1)$ and parallel to the $X$-axis is:

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