Find point on a curve that is part of a tangent line

Question

Undergrad Calculus III: I'm having trouble setting up for this question: There is a unique point P such that the tangent at P passes through the point $(-2,33,59)$. The curve is $r(t)=\langle 3+t, 1+2t^2, -3t-t^3\rangle$ Find $P$.

So far I think I need $r'(P)$ and $(-2,33,59)$ equated somehow, because I'm guessing the former is somehow our slope and the latter is a point, as needed to come up with an equation for a line. Then we will solve for the point $P$ by intersecting (equating) to the curve $r(t)$? Not sure if this is even the right approach, plz explain from basics if possible.

Answer 1

Hint. The direction of the tangent to the point at $t$ is given by $r'(t)=\langle 1,4t,-3-3t^2 \rangle$. So for some $s$, $$ r(t)+s\,r'(t)=X $$ where $X$ is the point you need it to pass through. So just solve for $t$ and $s$.

Answer 2

Given the point $(3+t, 1+2t^2, -3t-t^3)$, its derivative at $t$ is $(t, 4t, -3-3t^2)$. Match the $y$-slope as follows,

$$\frac{1+2t^2-33}{3+t-(-2)}=4t\tag{1}$$

which gives two solutions $t=-3$ and $t=-8$. Then, check which one satisfies the $z$-slope,

$$\frac{-3-t^3 -59}{3+t-(-2)}=-3-3t^2\tag{2}$$

The common solution is $t=-2$. Thus, the point on the curve is $(1,9,14)$.