Step 1: Tangency condition:
For $x^2/A-y^2/B=1$ with $A=B=7$, a line $lx+my=n$ touches it when $Al^2-Bm^2=n^2$.
Step 2: Test:
$7(16)-7(9)=49=7^2$. So the line is a tangent.
Step 3: Point of contact:
The point is $\left(\dfrac{Al}{n},\dfrac{-Bm}{n}\right)=\left(\dfrac{7\cdot4}{7},\dfrac{-7\cdot3}{7}\right)=(4,-3)$.
Step 4: Result:
Sum $=1$, option (C).
Final Answer:
The contact point (4, -3) gives sum 1.
\[ \boxed{C} \]