1. Interpret the given line

The line $\frac{x}{1}=\frac{y-1}{2}=\frac{z-2}{3}=t$ can be written in parametric form as $x=t,\qquad y=1+2t,\qquad z=2+3t.$ So a point on the line is $P_0=(0,1,2),$ and its direction vector is $\vec d=(1,2,3).$

2. Check Statement-2: Does the line bisect the segment joining $A$ and $B$?

Given $A=(1,0,7),\qquad B=(1,6,3).$ The midpoint of $AB$ is $M=\left(\frac{1+1}{2},\frac{0+6}{2},\frac{7+3}{2}\right)=(1,3,5).$

Now check whether $M$ lies on the line.

If $x=t$, then from $x=1$, we get $t=1.$ Then $y=1+2(1)=3,\qquad z=2+3(1)=5.$ So indeed, $M=(1,3,5)$ is on the given line.

Hence, the line bisects the segment $AB$.

So Statement-2 is true.

3. Check Statement-1: Is $A$ the mirror image of $B$ in the given line?

For a point to be the mirror image of another point in a line in 3D, the line must:

1. pass through the midpoint of the two points, and
2. be perpendicular to the segment joining them.

We already verified condition 1.

Now compute $\vec{AB}=B-A=(1-1,6-0,3-7)=(0,6,-4).$

Direction vector of the line is $\vec d=(1,2,3).$

Check perpendicularity using dot product: $\vec{AB}\cdot \vec d=(0)(1)+(6)(2)+(-4)(3)=0+12-12=0.$ Thus, $\vec{AB} \perp \vec d.$

Therefore the given line is the perpendicular bisector axis of segment $AB$, so reflection of $B$ in this line is indeed $A$.

Hence, Statement-1 is true.

4. Check whether Statement-2 explains Statement-1

Statement-2 says only that the line bisects the segment $AB$.

But for mirror image in a line, merely bisecting the segment is not sufficient. The line must also be perpendicular to the segment. Since Statement-2 does not include this perpendicularity condition, it is not a complete explanation of Statement-1.

Thus:

• Statement-1 is true
• Statement-2 is true
• Statement-2 is not the correct explanation of Statement-1

5. Correct option

Therefore, the correct answer is $\boxed{\text{A}}.$