For the given logic circuit, which of the following inputs...

Let the output of the OR gate be $Y_1$.

$Y_1 = A + B$

For LED-1 to glow, $Y_1$ must be $1$, which means $A + B = 1$.

Let the output of the first AND gate be $Y_2$.

$Y_2 = Y_1 \cdot C = (A + B) \cdot C$

Let the output of the second AND gate be $Y_3$.

$Y_3 = A \cdot Y_2 = A \cdot (A + B) \cdot C$

For LED-2 to glow, $Y_3$ must be $1$.

Simplifying the Boolean expression for $Y_3$:

$Y_3 = (A \cdot A + A \cdot B) \cdot C$

$Y_3 = (A + AB) \cdot C$

$Y_3 = A(1 + B) \cdot C$

$Y_3 = A \cdot C$

For $Y_3 = 1$, we require $A = 1$ and $C = 1$.

If $A = 1$, then $Y_1 = 1 + B = 1$ for any value of $B$, satisfying the condition for LED-1 to glow.

From the given options, the combination $A = 1, B = 0, C = 1$ satisfies these conditions.

Answer: $A = 1, B = 0, C = 1$