1. Find the intersection point of the given lines

The lines are $\frac{x}{3}+\frac{y}{4}=1 \quad \text{and} \quad \frac{x}{4}+\frac{y}{3}=1.$

Multiply by $12$: $4x+3y=12 \quad ...(1)$ $3x+4y=12 \quad ...(2)$

Subtracting, $(4x+3y)-(3x+4y)=0 \Rightarrow x-y=0 \Rightarrow x=y.$

Putting $x=y$ in (1): $4x+3x=12 \Rightarrow 7x=12 \Rightarrow x=y=\frac{12}{7}.$

So the variable line passes through $P\left(\frac{12}{7},\frac{12}{7}\right).$

2. Equation of a variable line meeting the axes at $A$ and $B$

Let the line cut the $x$-axis at $A(a,0)$ and the $y$-axis at $B(0,b)$. Then its intercept form is $\frac{x}{a}+\frac{y}{b}=1.$

Since it passes through $P\left(\frac{12}{7},\frac{12}{7}\right)$, $\frac{12/7}{a}+\frac{12/7}{b}=1.$

So, $\frac{12}{7}\left(\frac{1}{a}+\frac{1}{b}\right)=1$ $\frac{1}{a}+\frac{1}{b}=\frac{7}{12}.$

Thus, $\frac{a+b}{ab}=\frac{7}{12}. \quad ...(3)$

3. Coordinates of the midpoint of $AB$

Midpoint of $A(a,0)$ and $B(0,b)$ is $M\left(\frac{a}{2},\frac{b}{2}\right).$

Let the midpoint be $(x,y)$. Then $a=2x, \qquad b=2y.$

Substitute in (3): $\frac{2x+2y}{(2x)(2y)}=\frac{7}{12}.$

Simplify: $\frac{2(x+y)}{4xy}=\frac{7}{12}$ $\frac{x+y}{2xy}=\frac{7}{12}.$

Cross-multiplying, $12(x+y)=14xy$ $6(x+y)=7xy.$

Therefore the locus is $\boxed{7xy=6(x+y)}.$

4. Match with options

This corresponds to: $\boxed{\text{Option C}}$