Step 1: Understand the Condition

For a number to be divisible by both 7 and 11, it must be divisible by their least common multiple (LCM). Since 7 and 11 are prime numbers, their LCM is $7 \times 11 = 77$. Thus, we need to find which option is divisible by 77.

$\text{LCM}(7, 11) = 77$

Step 2: Check Option A ($3014$)

Divide 3014 by 7: $3014 \div 7 = 430.57$ (not an integer). Therefore, 3014 is not divisible by 7.

$3014 \div 7 = 430.57$

Step 3: Check Option B ($2534$)

Divide 2534 by 7: $2534 \div 7 = 362$ remainder 0? Let's check: $362 \times 7 = 2534$, wait: $362 \times 7 = 2534$? No, $362 \times 7 = 2534$ is incorrect ($362 \times 7 = 2534$ gives $362 \times 7 = 2534$). Let's check division: $2534 \div 7 = 362.0$ wait: $362 \times 7 = 2534$? Let's test $2534 / 7$: $7 \times 300 = 2100$, $434 / 7 = 62$, so $362$. Wait, $362 \times 7 = 2534$. But for 11: alternating sum $(4+5) - (3+2) = 4$, which is not divisible by 11.

$2534 \div 11 \implies \text{Remainder } 4$

Step 4: Check Option C ($2684$)

Check divisibility by 11: alternating sum $(4+6) - (8+2) = 0$ (divisible by 11). Now check divisibility by 7: $2684 \div 7 = 383.428$ (not divisible by 7).

$2684 \div 7 = 383.43$

Step 5: Check Option D ($1771$)

Check divisibility by 11: alternating sum $(1