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It is given $a=\begin{pmatrix} y-z \\ 3x \\ 2x \end{pmatrix} $ and points $A=(3,1,-2)$ and $D=(4,2,1)$. Give parameterization $t\in \left[0,1\right]$

$x=?$ Solution: $t+3$

$y=?$ Solution: $t+1$

$z=?$ Solution: $3t-2$

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    could you explain a bit more? It is hard to understand what you are trying to say. – 2017-02-20
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    Why don't you answer to the question of Arnaldo ? We do not understand what you mean ! – 2017-02-20
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    @JeanMarie I am translating it from another language and I dont know how to translate it better. – 2017-02-20

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If you're trying to find a parametrization between $A$ and $D$, we can create ${\bf v} = \vec{D} - \vec{A}= (4,2,1) - (3,1,-2) = (1,1,3)$. So, if we want to make a line in $3D$ passing through $A$ and $D$, we need the vector parallel to the line and an initial point. Let's take point $A$.

Then, $$\begin{align}\vec{r}(t) & = \vec{A} + t{\bf v} \\ &= (3,1,-2) + t(1,1,3) \\ &=(3 + t, 1 + t, -2+3t) \\&=(\underbrace{t+3}_{x(t)}, \underbrace{t+1}_{y(t)}, \underbrace{3t-2}_{z(t)}). \tag{Rearrange terms}\end{align}$$

Therefore, we get that $$\begin{align}a &= \pmatrix{y-z\\ 3x\\ 2x} \\&= \pmatrix{(t+1) - (3t-2)\\3(t+3)\\2(t+3)} \tag{Substitute in $x,y,z$}\\ &=\pmatrix{-2t+3 \\ 3t+9 \\ 2t+6} \tag{Combine like terms}\\ &= \pmatrix{-2\\3\\2}t + \pmatrix{3\\9\\6}\tag{Simplify}\end{align}$$