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(B) The equation of a plane passing through two points $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$ and perpendicular to a given plane $ax + by + cz = d$ c

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$(1, 1, 2)$

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(B) The equation of a plane passing through two points $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$ and perpendicular to a given plane $ax + by + cz = d$ can be found using the determinant form. The normal vector of the required plane is the cross product of the vector connecting the two points and the normal vector of the given plane.Let the points be $A(1, -1, 6)$ and $B(0, 0, 7)$. The vector $\vec{AB} = (0-1, 0-(-1), 7-6) = (-1, 1, 1)$.The normal vector of the plane $x - 2y + z = 6$ is $\vec{n_1} = (1, -2, 1)$.The normal vector of the required plane is $\vec{n} = \vec{AB} 	imes \vec{n_1} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ -1 & 1 & 1 \ 1 & -2 & 1 \end{vmatrix} = \hat{i}(1+2) - \hat{j}(-1-1) + \hat{k}(2-1) = 3\hat{i} + 2\hat{j} + 1\hat{k}$.The equation of the plane passing through $(0, 0, 7)$ with normal vector $(3, 2, 1)$ is $3(x-0) + 2(y-0) + 1(z-7) = 0$, which simplifies to $3x + 2y + z = 7$.Now, check the options by substituting the coordinates into the equation $3x + 2y + z = 7$:For $(1, 1, 2)$: $3(1) + 2(1) + 2 = 3 + 2 + 2 = 7$. This satisfies the equation.

$A$ point on the plane that passes through the points $(1, -1, 6)$, $(0, 0, 7)$ and is perpendicular to the plane $x - 2y + z = 6$ is

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