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(C) The equation of the $yz$-plane is $x = 0$. Since the required plane is perpendicular to the $yz$-plane,its normal vector must be perpendicular to

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$4y - z = 10$

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(C) The equation of the $yz$-plane is $x = 0$. Since the required plane is perpendicular to the $yz$-plane,its normal vector must be perpendicular to the normal vector of the $yz$-plane (which is $\hat{i} = (1, 0, 0)$). Let the points be $A(-1, 2, -2)$ and $B(-1, 3, 2)$. The vector $\vec{AB} = (-1 - (-1))\hat{i} + (3 - 2)\hat{j} + (2 - (-2))\hat{k} = 0\hat{i} + 1\hat{j} + 4\hat{k}$. Since the plane is perpendicular to the $yz$-plane,its normal vector $\vec{n}$ is perpendicular to $\hat{i} = (1, 0, 0)$ and $\vec{AB} = (0, 1, 4)$. Thus,$\vec{n} = \hat{i} 	imes \vec{AB} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & 0 & 0 \ 0 & 1 & 4 \end{vmatrix} = \hat{i}(0) - \hat{j}(4) + \hat{k}(1) = (0, -4, 1)$. The equation of the plane passing through $(-1, 2, -2)$ with normal vector $(0, -4, 1)$ is: $0(x + 1) - 4(y - 2) + 1(z + 2) = 0$ $-4y + 8 + z + 2 = 0$ $-4y + z + 10 = 0$ $4y - z = 10$.

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

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