The value of $(1001)^3$ is

$74.01 \%$ of $1301 + 9.99 \%$ of $1901 = ?$

$\sqrt{2498} \times \sqrt{626} \div \sqrt{99} = ?$

$3 \frac{6}{7} + 6 \frac{1}{4} - 5 \frac{1}{3} + 2 \frac{1}{2} = ?$

$\frac{16}{24} + \frac{4}{10} - \frac{1}{6} = ?$

Sum of three numbers is $132$. The first number is twice the second,and the third number is one-third of the first. Find the second number.