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Step-by-Step Detailed Solution
1. Write the given line in parametric form
The line is given by the symmetric equations
$ \frac{x}{1} = \frac{y - 1}{2} = \frac{z - 2}{3}. $
Let the common parameter be $t.$ Then:
$x = 1 \cdot t = t,$
$y - 1 = 2t \implies y = 1 + 2t,$
$z - 2 = 3t \implies z = 2 + 3t.$
Thus, the parametric form is
$$(x, y, z) = (0, 1, 2) + t\, (1, 2, 3).$$
2. Reflection of the point $(1,0,7)$ about this line
Let $P(1,0,7)$ be the point we want to reflect, and let the line pass through
$A(0,1,2)$ with direction vector $ \vec{d} = (1, 2, 3).$
2.1 Find the foot of the perpendicular from $P$ to the line
Suppose $F$ is the foot of the perpendicular from $P$ onto the line. Then
$$F = A + \lambda\,\vec{d} = (0 + \lambda,\, 1 + 2\lambda,\, 2 + 3\lambda).$$
The vector $ \overrightarrow{PF} $ must be perpendicular to $ \vec{d},$ so
$$ \overrightarrow{PF} \cdot \vec{d} = 0.$$
Explicitly, $ \overrightarrow{PF} = \bigl(\lambda - 1,\; (1+2\lambda) - 0,\; (2+3\lambda) - 7\bigr) = (\lambda - 1,\, 1 + 2\lambda,\, -5 + 3\lambda).$
Then
$$
(\lambda - 1)\cdot 1 + (1 + 2\lambda)\cdot 2 + (-5 + 3\lambda)\cdot 3 = 0.
$$
2.2 Solve for $ \lambda $
The equation becomes
$$
(\lambda - 1)
\;+\; 2(1 + 2\lambda)
\;+\; 3(-5 + 3\lambda)
= 0.
$$
Simplifying:
$$
\lambda - 1 + 2 + 4\lambda + (-15 + 9\lambda) = 0
$$
$$
\lambda + 4\lambda + 9\lambda - 1 + 2 - 15 = 0
$$
$$
14\lambda - 14 = 0
\quad \Longrightarrow \quad
14\lambda = 14
\quad \Longrightarrow \quad
\lambda = 1.
$$
Therefore, $F = (1,\; 3,\; 5).$
2.3 Find the reflected point $P'(\alpha, \beta, \gamma)$
The reflection $P'$ of $P$ across the line satisfies
$$ F = \text{midpoint of } P \text{ and } P'. $$
Hence,
$$
P' = 2F - P.
$$
Since $F=(1,3,5)$ and $P=(1,0,7),$
$$
(\alpha,\beta,\gamma)
= \bigl(2\cdot 1 - 1,\; 2\cdot 3 - 0,\; 2\cdot 5 - 7\bigr)
= (1,\; 6,\; 3).
$$
Thus, the image of $(1,0,7)$ in the given line is $(\alpha, \beta, \gamma)=(1,6,3).$
3. Direction cosines of the new line through $(1,6,3)$
We want a line through $(1,6,3)$ making angles
$ \tfrac{2\pi}{3}$ with the $y$-axis,
$ \tfrac{3\pi}{4}$ with the $z$-axis,
and an acute angle with the $x$-axis.
3.1 Express direction cosines
Let the direction cosines of this line be $(l, m, n)$, with
\[
l = \cos(\theta_x), \quad m = \cos(\theta_y), \quad n = \cos(\theta_z).
\]
From the problem:
\[
m = \cos\Bigl(\tfrac{2\pi}{3}\Bigr) = -\tfrac{1}{2},
\quad
n = \cos\Bigl(\tfrac{3\pi}{4}\Bigr) = -\tfrac{\sqrt{2}}{2}.
\]
Because the angle with the $x$-axis is acute, $l>0.$
3.2 Apply the identity for direction cosines
We know $l^2 + m^2 + n^2 = 1.$ Substituting
$ m=-\tfrac{1}{2}$ and $ n=-\tfrac{\sqrt{2}}{2},$ we get
\[
l^2 + \left(-\tfrac{1}{2}\right)^2 + \left(-\tfrac{\sqrt{2}}{2}\right)^2
= l^2 + \tfrac{1}{4} + \tfrac{1}{2}
= 1.
\]
Hence
\[
l^2 + \tfrac{3}{4} = 1
\quad\Longrightarrow\quad
l^2 = \tfrac{1}{4}.
\]
Taking $l>0$ gives
\[
l = \tfrac{1}{2}.
\]
Thus, the direction cosines are
\[
\bigl(\tfrac{1}{2},\; -\tfrac{1}{2},\; -\tfrac{\sqrt{2}}{2}\bigr).
\]
Any direction ratios are proportional to these cosines, so a convenient set of direction ratios is
\[
(1,\; -1,\; -\sqrt{2}).
\]
4. Parametric form of the new line
The required line through $(1,6,3)$ with direction ratios $(1, -1, -\sqrt{2})$ can be written as
\[
(x, y, z)
= (1, 6, 3) + \lambda \,(1,\,-1,\,-\sqrt{2}).
\]
Equivalently, in parametric form:
\[
x = 1 + \lambda,\quad
y = 6 - \lambda,\quad
z = 3 - \sqrt{2}\,\lambda.
\]
5. Verify the given options
We need to identify which of the provided points lies on this newly defined line.
The correct answer given is $(3,4,3 - 2\sqrt{2}).$
5.1 Check the point $(3,4,3 - 2\sqrt{2})$
If we set $x = 3,$ then from $x = 1+\lambda,$ we get $\lambda = 2.$
Substituting $\lambda = 2$ into
$y = 6 - \lambda$ gives $y = 6 - 2 = 4,$
and into
$z = 3 - \sqrt{2}\,\lambda = 3 - 2\sqrt{2}.$
Thus, $(3,4,3 - 2\sqrt{2})$ indeed satisfies the parametric equations, confirming it lies on the required line.
This matches the given correct answer in the options.
Conclusion
Therefore, the point on the line passing through the reflection of $(1, 0, 7)$
and satisfying the given angular conditions is
$$\boxed{(3,\,4,\,3 - 2\sqrt{2}).}$$
