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

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Step 1: Rewrite the equation of the line in standard form

The given equation of the line is $\frac{2x-4}{2}=\frac{y}{2}=\frac{2-z}{3}$. We can rewrite it as $\frac{x-2}{1}=\frac{y}{2}=\frac{z-2}{-3}$.

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Step 2: Find the general point on the line

Let the general point on the line be $P(2+\lambda, 2\lambda, 2-3\lambda)$.

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Step 3: Find the direction ratios of the line joining the given point and the general point

Let $A(-1,5,2)$ be the given point. The direction ratios of the line $AP$ are $(2+\lambda - (-1), 2\lambda - 5, 2-3\lambda - 2) = (3+\lambda, 2\lambda - 5, -3\lambda)$.

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Step 4: Use the perpendicularity condition

Since $AP$ is perpendicular to the given line, the dot product of their direction ratios is zero. Therefore, $1(3+\lambda) + 2(2\lambda - 5) - 3(-3\lambda) = 0$.

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Step 5: Solve for lambda

Simplifying the equation, we get $3+\lambda + 4\lambda - 10 + 9\lambda = 0$, which gives $14\lambda - 7 = 0$. Thus, $\lambda = \frac{1}{2}$.

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Step 6: Find the coordinates of the foot of the perpendicular

Substituting $\lambda = \frac{1}{2}$ in the general point $P$, we get the coordinates of the foot of the perpendicular as $P(2+\frac{1}{2}, 2(\frac{1}{2}), 2-3(\frac{1}{2})) = P(\frac{5}{2}, 1, \frac{1}{2})$.

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Step 7: Find the image of the point

Let the image of the point $A(-1,5,2)$ be $B(x,y,z)$. Since $P$ is the midpoint of $AB$, we have $\frac{x-1}{2} = \frac{5}{2}$, $\frac{y+5}{2} = 1$, and $\frac{z+2}{2} = \frac{1}{2}$. Solving these equations, we get $x-1 = 5$, $y+5 = 2$, and $z+2 = 1$. Therefore, $x = 6$, $y = -3$, and $z = -1$. The image point is $B(6, -3, -1)$.

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Step 8: Calculate the distance between the point and its image

The distance between $A(-1,5,2)$ and $B(6,-3,-1)$ is given by $\sqrt{(6-(-1))^2 + (-3-5)^2 + (-1-2)^2} = \sqrt{7^2 + (-8)^2 + (-3)^2} = \sqrt{49 + 64 + 9} = \sqrt{122}$.

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\r\n Final Answer: Image: (6, -3, -1), Distance: $\sqrt{122}$<\/span>\r\n <\/p>\r\n <\/div>\r\n <\/div>

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