Two lines $$\frac{x-3}{1} = \frac{y+1}{3} = \frac{z-6}{-1}$$ and $$\frac{x+5}{7} = \frac{y-2}{-6} = \frac{z-3}{4}$$ intersect at the point R. The reflection of R in the xy-plane has coordinates:

We have the first line written in symmetric form as $$\frac{x-3}{1}=\frac{y+1}{3}=\frac{z-6}{-1}$$. The standard way to convert this to a parametric representation is to introduce a single parameter, say $$t$$, and equate each fraction to $$t$$. Doing so gives

$$\frac{x-3}{1}=t,\qquad\frac{y+1}{3}=t,\qquad\frac{z-6}{-1}=t.$$

Multiplying each equation by its denominator we obtain

$$x-3 = t,\qquad y+1 = 3t,\qquad z-6 = -t.$$

Hence the coordinates of any point on this first line may be written as

$$x = 3+t,\qquad y = -1+3t,\qquad z = 6-t.$$

Now we take the second line $$\frac{x+5}{7}=\frac{y-2}{-6}=\frac{z-3}{4}$$ and introduce a second parameter, say $$s$$, so that

$$\frac{x+5}{7}=s,\qquad\frac{y-2}{-6}=s,\qquad\frac{z-3}{4}=s.$$

Again multiplying through by denominators we find

$$x+5 = 7s,\qquad y-2 = -6s,\qquad z-3 = 4s,$$

or equivalently

$$x = -5+7s,\qquad y = 2-6s,\qquad z = 3+4s.$$

The problem states that the two lines intersect at a point $$R(x,y,z)$$. Therefore the coordinates obtained from the first line must equal the coordinates obtained from the second line for some values of $$t$$ and $$s$$. Writing down the coordinate‐wise equalities we get the three equations

$$3+t \;=\; -5+7s\quad\text{(x‐coordinate)},$$

$$-1+3t \;=\; 2-6s\quad\text{(y‐coordinate)},$$

$$6-t \;=\; 3+4s\quad\text{(z‐coordinate)}.$$

Starting with the x‐coordinate equation, we solve for $$t$$:

$$t = -5 + 7s - 3 = 7s - 8.$$

Substituting this value of $$t$$ into the y‐coordinate equation, we obtain

$$-1 + 3(7s - 8) = 2 - 6s.$$

Simplifying the left‐hand side first,

$$-1 + 21s - 24 = 21s - 25.$$

Hence the equation becomes

$$21s - 25 = 2 - 6s.$$

Bringing like terms together, we add $$6s$$ to both sides and simultaneously add $$25$$ to both sides:

$$21s + 6s = 2 + 25.$$

This gives

$$27s = 27,$$

so

$$s = 1.$$

Recalling that $$t = 7s - 8$$, we substitute $$s = 1$$ to find

$$t = 7(1) - 8 = -1.$$

It is always good practice to confirm that these $$t$$ and $$s$$ values satisfy the third (z‐coordinate) equation as well. Substituting $$t = -1$$ and $$s = 1$$ into

$$6 - t = 3 + 4s,$$

the left side becomes $$6 - (-1) = 7$$ and the right side becomes $$3 + 4(1) = 7$$, so the equality holds and our intersection point is indeed consistent.

Now we substitute $$t = -1$$ into the parametric form of the first line (though we could equally well use the second line) to obtain the coordinates of the intersection point $$R$$:

$$x = 3 + (-1) = 2,$$

$$y = -1 + 3(-1) = -1 - 3 = -4,$$

$$z = 6 - (-1) = 6 + 1 = 7.$$

Thus $$R(2,\,-4,\,7).$$

The question now asks for the reflection of this point in the $$xy$$‐plane. By definition, reflection in the $$xy$$‐plane (whose equation is $$z=0$$) leaves the $$x$$‐ and $$y$$‐coordinates unchanged while reversing the sign of the $$z$$‐coordinate. Therefore, if $$R(x,y,z)$$ then its reflection $$R'(x',y',z')$$ satisfies

$$x' = x,\qquad y' = y,\qquad z' = -z.$$

Applying this to $$R(2,-4,7)$$, we obtain

$$x' = 2,\qquad y' = -4,\qquad z' = -7.$$

Hence the reflected point has coordinates $$R'(2,-4,-7).$$

Looking at the options provided, we see that the point $$(2,-4,-7)$$ corresponds to Option A.

Hence, the correct answer is Option A.