3. Determine which of the numbers is greater: $11^{\lg 121}$ or $10 \cdot 10^{\lg ^{2}} 11+11$.
Answer: the first.

Solution. Since $11^{\lg 121}=\left(11^{\lg 11}\right)^{2}$, and $10 \cdot 10^{\lg ^{2} 11}+11=10 \cdot\left(10^{\lg 11}\right)^{\lg 11}+11=10 \cdot 11^{\lg 11}+11$, we need to compare the numbers $x^{2}$ and $10 x+11$, where $x=11^{\lg 11}$. The expression $x^{2}-10 x-11=(x-11)(x+1)$ at the point $x=11^{\lg 11}>11$ takes a positive value, so the first number is greater.

Answer $₹$ to variant 2: the first. $\quad$ Answer $₹$ to variant 3: the second.

Answer $k$ to variant 4: the second.