Find a vector parametrization Find a vector parametrization for the line that
satisfies the given conditions.
Passes through the point P(1,−1, 2) and is parallel to the
line r(t) = t(3i − j + k).
What I tried is to plug in P to rt
P(1,-1,2) = (3*1 -(-1)+2)= 6 
Am I on the right track?
 A: No, you're not on the right track.  Remember that $i$, $j$, and $k$ are your standard unit vectors -- they are not variables that you can plug a point into.  So don't do that.

What you should do is this:
The vector equation of a line always looks like $$r(t) = vt + b$$ where $v$ is a vector parallel to your line, $b$ is a point on your line, and $t$ is your variable.  So you just need to figure out what your $v$ and $b$ vectors need to be (and because you've already used $r$ and $t$, maybe just change those to $s$ and $u$ or something).
Does that help?
A: Equation of Line passess through the point $A(\vec{a})$ and parallel to $\vec{b}$ is $\vec{r} = \vec{a}+\lambda \vec{b}$
So Here $\vec{a} = \hat{i}-\hat{j}+2\hat{k}$ and $\vec{b} = 3\hat{i}-\hat{j}+\hat{k}$.
So Our equation of line is $$\vec{r} = \left(\hat{i}-\hat{j}+2\hat{k}\right)+\lambda(3\hat{i}-\hat{j}+\hat{k}) = (1+3\lambda)\hat{i}+(-1-\lambda)\hat{j}+(2+\lambda)\hat{k}$$
Now Put $\vec{r} = x\hat{i}+y\hat{j}+z\hat{k}$
So we get $$x\hat{i}+y\hat{j}+z\hat{k}  = (1+3\lambda)\hat{i}+(-1-\lambda)\hat{j}+(2+\lambda)\hat{k}$$
So $\displaystyle x = 1+3\lambda$ and $y=-1-\lambda$ and $z=2+\lambda$
So Parametric equation of line is $$\displaystyle \frac{x-1}{3} = \frac{1+y}{-1} = \frac{z-2}{1} = \lambda.$$
where $\lambda$ is Scalar.
