Problem 70

Since no three points are collinear, every two points must determine a distinct line. Thus, there are $\dbinom{12}{2} = \frac{12\cdot11}{2} = 66$ lines.
Therefore, the answer is $\fbox{(D) 66}$

1. To determine the number of lines formed by 12 points in a plane where no three points are collinear, we need to count the number of ways to choose 2 points out of the 12 points. This is because a line is uniquely determined by any two distinct points.

2. The number of ways to choose 2 points from 12 points is given by the binomial coefficient:
$$\binom{12}{2}$$

3. The binomial coefficient $\binom{n}{k}$ is calculated using the formula:
$$\binom{n}{k} = \frac{n!}{k!(n-k)!}$$

4. Substituting $n = 12$ and $k = 2$ into the formula, we get:
$$\binom{12}{2} = \frac{12!}{2!(12-2)!} = \frac{12!}{2! \cdot 10!}$$

5. Simplifying the factorials, we have:
$$\binom{12}{2} = \frac{12 \times 11 \times 10!}{2 \times 1 \times 10!} = \frac{12 \times 11}{2 \times 1} = \frac{132}{2} = 66$$

6. Therefore, the number of lines determined by 12 points in a plane, no three of which are collinear, is:
$$\boxed{66}$$