Normal equation of a line. 
Find the normal equation of the line that passes across the point
  $(0,−1)$ and is parallel to the line whose parametric equation is
  $(−1+t,2−3t)$, $t\in\mathbb R$.

Is my idea correct?
Something normal to a line means it is perpendicular to it. Then If I want the Normal equation to be parallel to the parametric equation. Then I must use the perpendicular components to the parametric line to calculate the normal. 
Then from my parametric I obtain the vector $(1,-3)$ which is my direction and I want something orthogonal to it. 
Therefore $(1,-3) \cdot (x,y) = 0 $
this then gives me the orthogonal vector $(3, 1)$. now that I have an orthogonal vector to the parametric equation I can use this vector to get the normal which will be parallel to the parametric at the point $(0,-1)$.
$$3(x-0) + 1(y-(-1)) = 0$$ 
$$3x + y + 1 = 0$$ 
$y = -3x -1$ is a line normal a line that passes across a point and is parallel to the parametric. 
 A: The line equations are: standard, slope-y-intercept, point-slope, two-points and parametric which results from vector equation $\vec{r}=\vec{r_0}+t\cdot \vec{v}$
There is no normal equation of a line. We can construct a vector equation of a line using the normal vector, but that is not known as a normal equation. We can also construct the equation of a line normal to another line as in perpendicular. So there is an ambiguity in the text of the problem.
The best I could understand the text is to to construct a line equation using the normal vector to a line parallel to another line described parametrically.
If that is the case, then, using the previous notations,
$\vec{v}=(1, -3); \vec{n} \cdot \vec{v}=0 \Rightarrow \vec{n}=(3,1); \vec{r_0}=(0,-1)$
The line equation using the normal vector becomes:
$(\vec{r}-\vec{r_0}) \cdot \vec{n}=0\Leftrightarrow (x-0,y+1)\cdot (3,1)=0$
This leads to your answer $3x+y+1=0$, however this line is not normal (perpendicular) but parallel to the line described parametrically.
