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(C) The equation of the line passing through $Q(2, -2, 1)$ and perpendicular to the plane $x - 2y + z - 1 = 0$ is given by $\frac{x - 2}{1} = \frac{y

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$19$

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(C) The equation of the line passing through $Q(2, -2, 1)$ and perpendicular to the plane $x - 2y + z - 1 = 0$ is given by $\frac{x - 2}{1} = \frac{y + 2}{-2} = \frac{z - 1}{1} = k$. Any point on this line is $(k + 2, -2k - 2, k + 1)$. Since this point lies on the plane,we have $(k + 2) - 2(-2k - 2) + (k + 1) = 1$. $k + 2 + 4k + 4 + k + 1 = 1 \Rightarrow 6k + 7 = 1 \Rightarrow 6k = -6 \Rightarrow k = -1$. Thus,the foot of the perpendicular $P(x_1, y_1, z_1)$ is $(1, 0, 0)$. So,$l = x_1 + y_1 + z_1 = 1 + 0 + 0 = 1$. The perpendicular distance $d$ from $Q(2, -2, 1)$ to the plane is $d = \frac{|2 - 2(-2) + 1 - 1|}{\sqrt{1^2 + (-2)^2 + 1^2}} = \frac{|2 + 4 + 1 - 1|}{\sqrt{6}} = \frac{6}{\sqrt{6}} = \sqrt{6}$. Therefore,$d^2 = 6$. Finally,$l + 3d^2 = 1 + 3(6) = 1 + 18 = 19$.

Let $P(x_1, y_1, z_1)$ be the foot of the perpendicular drawn from the point $Q(2, -2, 1)$ to the plane $x - 2y + z = 1$. If $d$ is the perpendicular distance from the point $Q$ to the plane and $l = x_1 + y_1 + z_1$,then the value of $l + 3d^2$ is:

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