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

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Step 1: Define the Line and Point

Let the given line be $L: \frac{x}{1}=\frac{y-1}{2}=\frac{z-2}{3}$. The direction ratios of the line are $(1, 2, 3)$. Let the given point be $P'(1, 0, 7)$, which is the image of point $P(x, y, z)$ with respect to the line $L$.

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Step 2: Find the Midpoint of PP'

Let $M$ be the midpoint of $PP'$. The coordinates of $M$ are given by $M\left(\frac{x+1}{2}, \frac{y+0}{2}, \frac{z+7}{2}\right)$. Since $M$ lies on the line $L$, it must satisfy the equation of the line. Therefore, we have\r\n$$\frac{\frac{x+1}{2}}{1} = \frac{\frac{y}{2}-1}{2} = \frac{\frac{z+7}{2}-2}{3}$$\r\n$$\frac{x+1}{2} = \frac{y-2}{4} = \frac{z+3}{6} = k \text{ (say)}$$

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Step 3: Express x, y, and z in terms of k

From the above equation, we can express $x, y, z$ in terms of $k$ as follows:\r\n$$x = 2k - 1$$\r\n$$y = 4k + 2$$\r\n$$z = 6k - 3$$

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Step 4: Use the Direction Ratios

The direction ratios of the line $PP'$ are $(x-1, y-0, z-7)$. Since the line $PP'$ is perpendicular to the line $L$, the dot product of their direction ratios must be zero.\r\n$$(x-1)(1) + (y-0)(2) + (z-7)(3) = 0$$\r\nSubstituting the values of $x, y, z$ in terms of $k$, we get\r\n$$(2k - 1 - 1)(1) + (4k + 2)(2) + (6k - 3 - 7)(3) = 0$$\r\n$$(2k - 2) + (8k + 4) + (18k - 30) = 0$$\r\n$$28k - 28 = 0$$\r\n$$k = 1$$

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Step 5: Find the Coordinates of P

Substituting $k = 1$ in the expressions for $x, y, z$, we get\r\n$$x = 2(1) - 1 = 1$$\r\n$$y = 4(1) + 2 = 6$$\r\n$$z = 6(1) - 3 = 3$$\r\nTherefore, the coordinates of point $P$ are $(1, 6, 3)$.

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\r\n Final Answer: (1, 6, 3)<\/span>\r\n <\/p>\r\n <\/div>\r\n <\/div>

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