Equation of a line with the given description Given a vector-valued function defined by $${\displaystyle \mathbf{r}(t)\,=\,{\begin{pmatrix}t^3+1\\t^3+1\\2t+1\end{pmatrix}}}$$
Let $\mathbb T$ denote the tangent to the curve at $A=(2,2,3)$.
Then find the equation of the line $\mathbb L$ passing through the point $u=(1,-1,2)$,parallel to the plane $2x+y+z=0$ which intersects the tangent line $\mathbb T$.
The equation of the line is in the form:
$$\mathbb L=u+\vec vs$$
Since the line is parallel to the plane,we conclude that the direction of the line is the same as the plane's,let $\vec n=(2,1,1)$ denote the normal to the plane,then $\vec n \times \vec v=(1,1,1)$,which implies:
$$(v_2-v_3,2v_3-v_1,v_1-2v_2)=1$$
So:
$$v_2-v_3=1 \tag{1}$$
$$2v_3-v_1=1$$
$$v_1-2v_2=1\tag{2}$$
moreover $\vec n \cdot \vec v=0$,which implies:
$$2v_1+v_2+v_3=0 \tag{3}$$
Substituting  $(1)$ and $(2)$ into $(3)$ follows:
$$v_1=2/3$$$$v_2=-1/6$$$$v_3=-7/6$$
So the equation of the line is :
$$\mathbb L=(1,-1,2)+ (\frac{2}{3},-\frac{1}{6},-\frac{7}{6})s$$
With the parametric equation :
$$x=\frac{2}{3}s+1$$
$$y=-\frac{1}{6}s-1$$
$$z=-\frac{7}{6}s+2$$
Since the line intersects the tangent line to the curve at a point with coordinate $(2,2,3)$,we see that $s=3/2$,however substituting this to the $y$ and $z$ we don't get $y=2$ and $z=3$ respectively,so where was I wrong?
 A: As stated, your problem can have no solution: if a line through $u=(1,-1,2)$ also passes through $A=(2,2,3)$, $\overrightarrow{Au}=(1,3,1)$ is a directing vector, and if the line is parallel to the plane with equation $2x+y+z=0$, with normal vector $\vec n=(2,1,1)$, one should have $$\overrightarrow{Au}\cdot \vec  n=0,$$
which is not the case.
That's why I think the real meaning of the question is that a line through $u$, parallel to the given  plane  meets the tangent drawn from point $A$ on the curve.
A: I'm not sure how you got $\vec{n}\times \vec{v} = (1,1,1)$ but it is wrong.
The correct condition that $\Bbb{L} = U+t\vec{v}$ is parallel to the plane $2x+y+z=0$ is that $\vec{v}$ is orthogonal to the normal vector of the plane $\vec{n} = (2,1,1)$, or
$$0 = \vec{n} \cdot \vec{v} = 2v_1+v_2+v_3.$$
The other condition is that $\mathbb{L}$ intersects the tangent $\mathbb{T}$ at the curve $\mathbf{r}$ at $A = (2,2,3) = \mathbf{r}(1)$. This tangent is given by
$$\mathbb{T}=A + s\mathbf{r}'(1) = (2,2,3)+s(3,3,2), \quad s \in \Bbb{R}.$$
$\mathbb{L}$ and $\mathbb{T}$ intersect so there are $s,t \in \Bbb{R}$ such that
\begin{align}
U+t\vec{v} = A+s(3,3,2) &\implies \vec{AU} + t\vec{v} - s(3,3,2) = 0 \\
&\implies \vec{AU}, \vec{v}, (3,3,2) \text{ are linearly dependent}
\end{align}
so with $\vec{AU} = (-1,-3,-1)$ we have
$$0 = \det(\vec{AU}, \vec{v}, (3,3,2)) = \begin{vmatrix} -1 & -3 & -1 \\ v_1 & v_2 & v_3 \\ 3 & 3 & 2 \end{vmatrix} = 3v_1+v_2-6v_3$$
and combining this with $2v_1+v_2+v_3 = 0$ we get $\vec{v} = (7,-15,1)$ up to multiplication by scalar. Therefore your line is given by
$$\Bbb{L} = U+t\vec{v} = (1,-1,2) + t(7,-15,1), \quad t \in \Bbb{R}.$$
