Select the cars one by one an for $i=1,2,3$ let $F_i$ denote the event that the car chosen as number $i$ is faulty. Then the event that exactly $2$ cars will be faulty is:$$E:=(F_1\cap F_2\cap F_3^{\complement})\cup(F_1\cap F_2^{\complement}\cap F_3)\cup(F_1^{\complement}\cap F_2\cap F_3)$$
This is a union of $3$ disjoint sets and we can calculate the probability of e.g. the first like this:$$P(F_1\cap F_2\cap F_3^{\complement})=P(F_1)P(F_2\mid F_1)P(F_3^{\complement}\mid F_1\cap F_2)=\frac3{100}\frac2{99}\frac{97}{98}$$
On a similar way the two others can be calculate and it will appear that we find the same outcome.
Then we find:$$P(E)=3\times\frac3{100}\frac2{99}\frac{97}{98}$$
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Making use of the theory concerning hypergeometric distribution we find the outcome:$$\frac{\binom32\binom{97}1}{\binom{100}3}$$