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(D) The equation of a plane parallel to $2x - y + 2z = 0$ is of the form $2x - y + 2z + k = 0$. Since this plane passes through the point $(-2, 1, -1)

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$(-1, 3, -1)$

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(D) The equation of a plane parallel to $2x - y + 2z = 0$ is of the form $2x - y + 2z + k = 0$. Since this plane passes through the point $(-2, 1, -1)$,we substitute these coordinates into the equation: $2(-2) - (1) + 2(-1) + k = 0 \Rightarrow -4 - 1 - 2 + k = 0 \Rightarrow k = 7$. Thus,the equation of the plane $\pi$ is $2x - y + 2z + 7 = 0$. Let $(a, b, c)$ be the foot of the perpendicular from the point $(1, 2, 1)$ to the plane $\pi$. The line passing through $(1, 2, 1)$ and perpendicular to the plane has the direction ratios $(2, -1, 2)$. Thus,the line equation is $\frac{a - 1}{2} = \frac{b - 2}{-1} = \frac{c - 1}{2} = \lambda$. This gives $a = 2\lambda + 1$,$b = -\lambda + 2$,and $c = 2\lambda + 1$. Since $(a, b, c)$ lies on the plane $\pi$,we substitute these into the plane equation: $2(2\lambda + 1) - (-\lambda + 2) + 2(2\lambda + 1) + 7 = 0 \Rightarrow 4\lambda + 2 + \lambda - 2 + 4\lambda + 2 + 7 = 0 \Rightarrow 9\lambda + 9 = 0 \Rightarrow \lambda = -1$. Substituting $\lambda = -1$ back into the expressions for $a, b, c$: $a = 2(-1) + 1 = -1$,$b = -(-1) + 2 = 3$,$c = 2(-1) + 1 = -1$. Therefore,the foot of the perpendicular is $(-1, 3, -1)$.

Let $\pi$ be the plane that passes through the point $(-2, 1, -1)$ and is parallel to the plane $2x - y + 2z = 0$. Then the foot of the perpendicular drawn from the point $(1, 2, 1)$ to the plane $\pi$ is

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