1. Let the required line cut the $x$-axis at $P(a,0)$ and the $y$-axis at $Q(0,b)$.

2. The intercept between the axes is the segment $PQ$. Since this segment is bisected at $A(3,4)$, the midpoint of $P(a,0)$ and $Q(0,b)$ is $(3,4)$.

3. Using the midpoint formula, $\left(\frac{a+0}{2},\frac{0+b}{2}\right)=(3,4)$ So, $\frac{a}{2}=3 \Rightarrow a=6$ $\frac{b}{2}=4 \Rightarrow b=8$

4. Hence the line has intercepts $6$ and $8$ on the axes. Its intercept form is $\frac{x}{6}+\frac{y}{8}=1$

5. Multiply by $24$: $4x+3y=24$

6. Now check the options:

• A: $x+y=7$ gives intercepts $7,7$; midpoint $(3.5,3.5)$, not $(3,4)$.
• B: $3x-4y+7=0$ does not give the required midpoint condition.
• C: $4x+3y=24$ gives intercepts $(6,0)$ and $(0,8)$; midpoint is $(3,4)$.
• D: $3x+4y=25$ gives intercepts $\left(\frac{25}{3},0\right)$ and $\left(0,\frac{25}{4}\right)$; midpoint is not $(3,4)$.

Therefore, the correct equation is $\boxed{4x+3y=24}$