The sum of squares of three natural numbers is 138 and sum of their products taken two at a time is What is the sum of these numbers?

The correct answer is (d).

Let $a$, $b$, and $c$ be the three natural numbers. We know that $a^2 + b^2 + c^2 = 138$ and $ab + ac + bc = 60$. We can use the Cauchy–Schwarz inequality to get a lower bound for the sum of the numbers:

$$(a + b + c)^2 \ge (a^2 + b^2 + c^2)(ab + ac + bc) = 138 \cdot 60 = 8280.$$

This means that $a + b + c \ge \sqrt{8280} \approx 91.0$.

On the other hand, we can use the triangle inequality to get an upper bound for the sum of the numbers:

$$a + b + c \le a^2 + b^2 + c^2 + 2(ab + ac + bc) = 138 + 2 \cdot 60 = 258.$$

Therefore, the sum of the numbers must be between 91 and 258. Since 10, 20, and 30 are all less than 91, they cannot be the sum of the numbers. Therefore, the correct answer is (d).