The correct answer is $\boxed{\text{D}}$.
The vector normal to the surface $2xz^2 – 3xy – 4x = 7$ at the point $(1, -1, 2)$ is the gradient of the surface evaluated at that point. The gradient of a surface is a vector that points in the direction of the greatest rate of change of the surface. The gradient of the surface $2xz^2 – 3xy – 4x = 7$ is
$$\nabla f = \begin{bmatrix} 4z – 3y \ \ 2x – 3z \ \ -4 \end{bmatrix}$$
Evaluating the gradient at the point $(1, -1, 2)$ gives
$$\nabla f(1, -1, 2) = \begin{bmatrix} 4 – 3 \ \ 2 – 3 \ \ -4 \end{bmatrix} = \boxed{7i – 5j + 8k}$$
Therefore, the vector that is normal to the surface $2xz^2 – 3xy – 4x = 7$ at the point $(1, -1, 2)$ is $\boxed{7i – 5j + 8k}$.
Here is a brief explanation of each option:
- Option A: $2i – 3j + 8k$ is not normal to the surface $2xz^2 – 3xy – 4x = 7$ at the point $(1, -1, 2)$. To see this, we can take the dot product of $2i – 3j + 8k$ with the gradient of the surface at the point $(1, -1, 2)$. The dot product of two vectors is zero if and only if the vectors are orthogonal. Therefore, if $2i – 3j + 8k$ were normal to the surface, the dot product would be zero. However, the dot product is
$$(2i – 3j + 8k) \cdot \nabla f(1, -1, 2) = 2 \cdot 4 – 3 \cdot 3 + 8 \cdot 8 = 41 > 0$$
Therefore, $2i – 3j + 8k$ is not normal to the surface.
Option B: $2i + 3j + 4k$ is not normal to the surface $2xz^2 – 3xy – 4x = 7$ at the point $(1, -1, 2)$. The same argument as for option A shows that $2i + 3j + 4k$ is not normal to the surface.
Option C: $7i – 3j + 8k$ is normal to the surface $2xz^2 – 3xy – 4x = 7$ at the point $(1, -1, 2)$. This is because the dot product of $7i – 3j + 8k$ with the gradient of the surface at the point $(1, -1, 2)$ is zero.
Option D: $7i – 5j + 8k$ is normal to the surface $2xz^2 – 3xy – 4x = 7$ at the point $(1, -1, 2)$. This is because the dot product of $7i – 5j + 8k$ with the gradient of the surface at the point $(1, -1, 2)$ is zero.