Hint:
Using your approach, note that ,if $Q=(x_Q,y_Q)$ is the point of tangency than the slope of the line is:
$$
m=\frac{y_Q-5}{x_Q-5}
$$
Since $Q$ is point of the parabola, we have $y_Q=x_Q^2-2x_Q-1$, and we want that this slope is the same as the derivative of the function at $x_Q$, so we have the equation:
$$
\frac{f(x_Q)-5}{x_Q-5}=f'(x_Q)
$$
that is:
$$
\frac{x_Q^2-2x_Q-1-5}{x_Q-5}=2x_Q-2
$$
solve this equation and find $x_Q$, so you can find the point $Q$ and the slope of the tangent.
But you can use also another approach that does not require derivatives.
The lines passing for the given point are:
$$
y-5=m(x-5)
$$
and we want the line that is tangent to the parabola, this means that this line has only one common point with the parabola, so, write the system:
$$
\begin {cases}
y-5=m(x-5)\\
y=x^2-2x-1
\end{cases}
$$
and find for what values of $m$ it has only one solution, i.e. for what $m$ the discriminant of the system is equal to 0. This is the slope of the tangent.
Let $\vec{r}(t) = \langle 2−t,−1−t^2,−2t−3t^3\rangle$, where $t\in \mathbb{R}$. Then $\vec{r}'(t) = \langle -1, -2t, -2 -9t^2\rangle$.
For any $t_0$ in the domain, a parametric equation of the tangent line at $t_0$ is given by
$$
\vec{T}(t) = t \:\vec{r}'(t_0) + \vec{r}(t_0), \mbox{ where } t\in \mathbb{R}.
$$
Now, we want to find an equation of the tangent line when it is passing through $(2,8, -162)$.
Thus, setting the following equal to each other,
\begin{align*}
t\langle -1, -2t, -2 -9t^2\rangle + \langle 2−t_0,−1−t_0^2,−2t_0−3t_0^3\rangle = \langle 2,8, -162 \rangle,
\end{align*}
we need to solve for $t_0$.
The above vector equation gives us three equations:
\begin{align*}
-t+2-t_0&=2, \hspace{4mm} (\dagger) \\
-2t_0t-1-t_0^2&=8, \hspace{4mm} (\ddagger) \\
-2t-9t_0^2t-2t_0-3t_0^3&=-162. \hspace{4mm} (\Omega)\\
\end{align*}
Use the first equation $(\dagger)$ to solve for $t:$
$$
t=-t_0.
$$
Let's substitute this into the second equation $(\ddagger)$:
$$
2t_0^2 -1-t_0^2 = 8,
$$
which gives us $t_0^2 = 9$.
So $t_0 = \pm 3$ while $t=\mp 3$.
If $t_0=3$, then $t=-3$. We substitute these into the third equation $(\Omega)$ to obtain:
$$
6+9\cdot 9\cdot 3 -2(3)-3(27) = 9(27)-3(27)=6(27) \not= -162.
$$
We conclude that $t_0=-3$ while $t=3$, which we can check using the third equation $(\Omega)$:
$$
-6-9\cdot 9\cdot 3 +2(3)+3(27) = -9(27)+3(27)=-6(27) = -162.
$$
So $P$ has position vector $\vec{r}(-3)=\langle 5,-10,87 \rangle $, which is when its tangent line passes through the point $(2,8,−162)$.
Best Answer
Two surfaces, paraboloid $z=x^2+y^2$ and plane $z=3-x$, intersect on ellipse: $$x^2+y^2=3-x \iff x^2+x+y^2=3 \iff (x+\frac{1}{2})^2-\frac{1}{4}+y^2 = 3 \iff (x+\frac{1}{2})^2+y^2=\frac{13}{4} \\$$
$$\iff \frac{(x+\frac{1}{2})^2}{\frac{13}{4}}+\frac{y^2}{\frac{13}{4}}=1$$
Now the tangent on the intersection curve at the point $P=(1,1,2)$ is easy to find, as well as it is easy to check which one of 4 points lie on this tangent...
Figure.