Detailed Solution<\/h3>\r\n <\/div>\r\n\r\n

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Step 1: Convert the line equation to standard form

The given line equation is $\frac{x}{1}=\frac{1-y}{-2}=\frac{2z-4}{6}$. We can rewrite this as $\frac{x}{1}=\frac{y-1}{2}=\frac{z-2}{3}$. This gives us the direction ratios of the line as $(1, 2, 3)$.

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Step 2: Find the equation of the plane

The line is normal to the plane, so the direction ratios of the line are the direction ratios of the normal to the plane. Thus, the equation of the plane is of the form $x + 2y + 3z = d$. Since the line passes through the point $(0, 1, 2)$, this point also lies on the plane. Substituting this point into the equation of the plane, we get $0 + 2(1) + 3(2) = d$, which gives $d = 8$. Therefore, the equation of the plane is $x + 2y + 3z = 8$.

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Step 3: Find the coordinates of the foot of the perpendicular from P to the plane

Let the coordinates of the foot of the perpendicular from $P(1, 6, 3)$ to the plane $x + 2y + 3z = 8$ be $Q(x_1, y_1, z_1)$. The direction ratios of the line $PQ$ are $(x_1 - 1, y_1 - 6, z_1 - 3)$. Since $PQ$ is normal to the plane, the direction ratios of $PQ$ are proportional to the direction ratios of the normal to the plane. Thus, we have $\frac{x_1 - 1}{1} = \frac{y_1 - 6}{2} = \frac{z_1 - 3}{3} = k$. This gives $x_1 = k + 1$, $y_1 = 2k + 6$, and $z_1 = 3k + 3$. Since $Q$ lies on the plane, we have $(k + 1) + 2(2k + 6) + 3(3k + 3) = 8$. Simplifying, we get $k + 1 + 4k + 12 + 9k + 9 = 8$, which gives $14k + 22 = 8$, so $14k = -14$, and $k = -1$. Thus, $x_1 = -1 + 1 = 0$, $y_1 = 2(-1) + 6 = 4$, and $z_1 = 3(-1) + 3 = 0$. Therefore, the coordinates of the foot of the perpendicular are $Q(0, 4, 0)$.

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Step 4: Find the coordinates of the image point

Let the image of $P(1, 6, 3)$ be $R(x_2, y_2, z_2)$. Since $Q$ is the midpoint of $PR$, we have $x_1 = \frac{x_2 + 1}{2}$, $y_1 = \frac{y_2 + 6}{2}$, and $z_1 = \frac{z_2 + 3}{2}$. Substituting the coordinates of $Q$, we get $0 = \frac{x_2 + 1}{2}$, $4 = \frac{y_2 + 6}{2}$, and $0 = \frac{z_2 + 3}{2}$. This gives $x_2 = -1$, $y_2 = 8 - 6 = 2$, and $z_2 = -3$. Therefore, the coordinates of the image point are $R(-1, 2, -3)$.

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Step 5: Find the distance between P and its image

The distance between $P(1, 6, 3)$ and $R(-1, 2, -3)$ is given by $\sqrt{(-1 - 1)^2 + (2 - 6)^2 + (-3 - 3)^2} = \sqrt{(-2)^2 + (-4)^2 + (-6)^2} = \sqrt{4 + 16 + 36} = \sqrt{56} = 2\sqrt{14}$.

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\r\n Final Answer: The coordinates of the image point are $(-1, 2, -3)$ and the distance between the point P and its image is $2\sqrt{14}$.<\/span>\r\n <\/p>\r\n <\/div>\r\n <\/div>

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