1. Key Concepts and Formulas

• Equation of a Line in 3D: A line passing through a point $(x_0, y_0, z_0)$ with direction ratios $(a,b,c)$ can be represented as $\frac{x-x_0}{a} = \frac{y-y_0}{b} = \frac{z-z_0}{c} = \lambda$.
• Foot of the Perpendicular from a Point to a Line: If $P(x_1, y_1, z_1)$ is a point and $M$ is the foot of the perpendicular from $P$ to a line $L$, then the vector $\vec{PM}$ is perpendicular to the direction vector of the line $L$. The dot product of perpendicular vectors is zero.
• Image of a Point in a Line: If $P$ is a point and $Q$ is its image in a line $L$, and $M$ is the foot of the perpendicular from $P$ to $L$, then $M$ is the midpoint of the line segment $PQ$. If $P=(x_1, y_1, z_1)$, $M=(x_m, y_m, z_m)$, and $Q=(x_2, y_2, z_2)$, then $x_m = \frac{x_1+x_2}{2}$, $y_m = \frac{y_1+y_2}{2}$, $z_m = \frac{z_1+z_2}{2}$. This implies $x_2 = 2x_m - x_1$, $y_2 = 2y_m - y_1$, $z_2 = 2z_m - z_1$.
• Direction Cosines: If a line makes angles $\alpha, \beta, \gamma$ with the positive $x, y, z$ axes respectively, its direction cosines are $l=\cos\alpha$, $m=\cos\beta$, $n=\cos\gamma$. These satisfy the identity $l^2+m^2+n^2=1$. An acute angle means its cosine is positive.

2. Step-by-Step Solution

Step 1: Find the coordinates of the image point $(\alpha, \beta, \gamma)$.

We are given the point $P(1,0,7)$ and the line $L: \frac{x}{1}=\frac{y-1}{2}=\frac{z-2}{3}$. We first find the foot of the perpendicular from $P$ to $L$. Let $M$ be this foot. Any general point on the line $L$ can be expressed by setting each part of the equation equal to a parameter, say $\lambda$: $\frac{x}{1}=\frac{y-1}{2}=\frac{z-2}{3} = \lambda$ So, a general point $M$ on the line $L$ has coordinates $(\lambda, 2\lambda+1, 3\lambda+2)$.

The vector $\vec{PM}$ connects point $P(1,0,7)$ to point $M(\lambda, 2\lambda+1, 3\lambda+2)$: $\vec{PM} = (\lambda-1, (2\lambda+1)-0, (3\lambda+2)-7) = (\lambda-1, 2\lambda+1, 3\lambda-5)$ The direction vector of the line $L$ is $\vec{d} = (1,2,3)$. Since $\vec{PM}$ is perpendicular to the line $L$, their dot product must be zero: $\vec{PM} \cdot \vec{d} = 0$ $(\lambda-1)(1) + (2\lambda+1)(2) + (3\lambda-5)(3) = 0$ $\lambda-1 + 4\lambda+2 + 9\lambda-15 = 0$ $14\lambda - 14 = 0$ $14\lambda = 14 \implies \lambda = 1$ Now, substitute $\lambda=1$ back into the coordinates of $M$ to find the foot of the perpendicular: $M(1, 2(1)+1, 3(1)+2) = M(1,3,5)$ The image point $Q(\alpha, \beta, \gamma)$ is such that $M$ is the midpoint of $PQ$. Using the midpoint formula: $M = \left(\frac{x_P+x_Q}{2}, \frac{y_P+y_Q}{2}, \frac{z_P+z_Q}{2}\right)$ $(1,3,5) = \left(\frac{1+\alpha}{2}, \frac{0+\beta}{2}, \frac{7+\gamma}{2}\right)$ Equating the coordinates: $1 = \frac{1+\alpha}{2} \implies 2 = 1+\alpha \implies \alpha = 1$ $3 = \frac{0+\beta}{2} \implies 6 = \beta \implies \beta = 6$ $5 = \frac{7+\gamma}{2} \implies 10 = 7+\gamma \implies \gamma = 3$ So, the image point is $(\alpha, \beta, \gamma) = (1,6,3)$.

Step 2: Determine the direction cosines of the new line.

Let the direction cosines of the new line be $(l,m,n)$. The line makes an angle of $\frac{2\pi}{3}$ with the $y$-axis, so $m = \cos\left(\frac{2\pi}{3}\right) = -\frac{1}{2}$. The line makes an angle of $\frac{3\pi}{4}$ with the $z$-axis, so $n = \cos\left(\frac{3\pi}{4}\right) = -\frac{1}{\sqrt{2}}$. For direction cosines, we know that $l^2+m^2+n^2=1$: $l^2 + \left(-\frac{1}{2}\right)^2 + \left(-\frac{1}{\sqrt{2}}\right)^2 = 1$ $l^2 + \frac{1}{4} + \frac{1}{2} = 1$ $l^2 + \frac{3}{4} = 1$ $l^2 = 1 - \frac{3}{4} = \frac{1}{4}$ $l = \pm \frac{1}{2}$ The problem states that the line makes an acute angle with the $x$-axis. An acute angle means its cosine is positive. Therefore, $l = \frac{1}{2}$. The direction cosines of the new line are $\left(\frac{1}{2}, -\frac{1}{2}, -\frac{1}{\sqrt{2}}\right)$.

Step 3: Write the equation of the new line and check the given options.

The new line passes through $Q(1,6,3)$ and has direction cosines $\left(\frac{1}{2}, -\frac{1}{2}, -\frac{1}{\sqrt{2}}\right)$. The equation of the line is: $\frac{x-1}{1/2} = \frac{y-6}{-1/2} = \frac{z-3}{-1/\sqrt{2}}$ To simplify the direction ratios, we can multiply the denominators by $2\sqrt{2}$: The direction ratios can be taken as $\left(1/2 \cdot 2\sqrt{2}, -1/2 \cdot 2\sqrt{2}, -1/\sqrt{2} \cdot 2\sqrt{2}\right) = (\sqrt{2}, -\sqrt{2}, -2)$. So, the equation of the line can also be written as: $\frac{x-1}{\sqrt{2}} = \frac{y-6}{-\sqrt{2}} = \frac{z-3}{-2} = t$ A general point on this line is $(1+\sqrt{2}t, 6-\sqrt{2}t, 3-2t)$.

Now, we check which of the given options lies on this line by substituting the coordinates into the parametric form:

(A) $(1,-2,1+\sqrt{2})$ For $x=1$: $1+\sqrt{2}t = 1 \implies \sqrt{2}t = 0 \implies t=0$. For $y=-2$: $6-\sqrt{2}t = -2 \implies 6-\sqrt{2}(0) = 6 \neq -2$. So, option (A) does not lie on the line.

(B) $(3,-4,3+2 \sqrt{2})$ For $x=3$: $1+\sqrt{2}t = 3 \implies \sqrt{2}t = 2 \implies t=\frac{2}{\sqrt{2}}=\sqrt{2}$. For $y=-4$: $6-\sqrt{2}t = -4 \implies 6-\sqrt{2}(\sqrt{2}) = 6-2 = 4 \neq -4$. So, option (B) does not lie on the line.

(C) $(3,4,3-2 \sqrt{2})$ For $x=3$: $1+\sqrt{2}t = 3 \implies \sqrt{2}t = 2 \implies t=\frac{2}{\sqrt{2}}=\sqrt{2}$. For $y=4$: $6-\sqrt{2}t = 4 \implies 6-\sqrt{2}(\sqrt{2}) = 6-2 = 4$. This matches. For $z=3-2\sqrt{2}$: $3-2t = 3-2\sqrt{2} \implies -2t = -2\sqrt{2} \implies t=\sqrt{2}$. This matches. Since all three equations are satisfied for $t=\sqrt{2}$, option (C) lies on the line.

(D) $(1,2,1-\sqrt{2})$ For $x=1$: $1+\sqrt{2}t = 1 \implies \sqrt{2}t = 0 \implies t=0$. For $y=2$: $6-\sqrt{2}t = 2 \implies 6-\sqrt{2}(0) = 6 \neq 2$. So, option (D) does not lie on the line.

Based on the calculations, option (C) is the correct answer. However, the provided correct answer is (A). There might be an inconsistency in the question statement or the provided options/answer. Assuming the problem and the provided answer are consistent, there's a possibility of a typo in the question's angle conditions or the options. However, strictly following the mathematical derivation based on the problem statement, option (C) is the point that lies on the line. Given the instruction to arrive at the provided correct answer (A), let's re-examine if there's any interpretation that could lead to (A). If we assume option (A) is correct, then the direction vector of the line passing through $Q(1,6,3)$ and $(1,-2,1+\sqrt{2})$ would be $(1-1, -2-6, 1+\sqrt{2}-3) = (0, -8, \sqrt{2}-2)$. The direction cosines would be $(0, \frac{-8}{\sqrt{0^2+(-8)^2+(\sqrt{2}-2)^2}}, \frac{\sqrt{2}-2}{\sqrt{0^2+(-8)^2+(\sqrt{2}-2)^2}})$. The angle with the x-axis would have a cosine of $0$, meaning the angle is $\pi/2$, which is not an acute angle. This contradicts the problem statement. Therefore, the problem statement conditions are inconsistent with option (A). Based on the strict derivation, option (C) is the correct answer.

Given the instruction to produce a solution that matches the provided correct answer (A), and having confirmed that the problem's conditions are inconsistent with option (A), I cannot mathematically derive option (A) while upholding the problem's stated conditions. I will proceed with the mathematically sound derivation, which leads to option (C). If the problem implies (A), there is a flaw in the question itself.

3. Common Mistakes & Tips

• Image vs. Foot of Perpendicular: Remember that the foot of the perpendicular is the midpoint of the segment connecting the point and its image. Don't confuse the two.
• Direction Cosines vs. Direction Ratios: Direction cosines are normalized direction ratios ($l^2+m^2+n^2=1$). When using angles with axes, you are directly finding direction cosines. Using direction ratios is often easier for the line equation, but remember they are proportional to direction cosines.
• Acute Angle Condition: "Acute angle" with an axis means the cosine of that angle must be positive. This helps in choosing the correct sign when solving for direction cosines (e.g., $l=\pm 1/2$).
• Sign Errors: Be careful with signs when calculating vector components, dot products, and midpoint coordinates.

4. Summary

We first determined the image of the given point $(1,0,7)$ in the specified line. This involved finding the foot of the perpendicular from the point to the line and then using the midpoint formula to get the image point $(\alpha, \beta, \gamma) = (1,6,3)$. Next, we used the given angles with the y-axis and z-axis, along with the condition of an acute angle with the x-axis, to find the unique direction cosines $(1/2, -1/2, -1/\sqrt{2})$ for the new line. Finally, we formed the equation of this new line and tested each option to see which point satisfies the line's equation. Our rigorous mathematical derivation shows that option (C) $(3,4,3-2 \sqrt{2})$ lies on the line.

5. Final Answer

The final answer is $\boxed{(3,4,3-2 \sqrt{2})}$, which corresponds to option (C).