Chapter 13 - Section 13.1 - Vector Functions and Space Curves - 13.1 Exercises - Page 855: 49

Yes

We need to rearrange the equation as: $t^2-4t+3=0$ $t^2--3t-t+3=0$ or, $ t= 3$ Now, we need to rearrange the equation as: $t^2-7t+12=0$ $t^2-6t-t+12=0$ This gives: $ t= 3$ Further, set the $z$ component of the two equations equal to each other first. We have $t^2=5t-6$ or, $t^2-5t+6=0 \implies t^2-6t+t+6=0$ This gives $t=3$ Hence, all three equations gives the same result. So, the answer is Yes.