Finding parametric equations given a point, orthogonal to another vector, and contained in a plane 
Find parametric equations for the line with the following properties. The line passes through the origin, it is contained in the plane $x-2y+z=0$, and is orthogonal to the vector $v = \langle3,4,2\rangle$.

I know that the vector is orthogonal to $v$ when the dot product of the vector I am looking for and $v$ is equal to $0$. But, how would I make sure that said vector is also meeting the other requirements? 
 A: The parametric equations will be of the form
$$x=0+at $$
$$y=0+bt $$
$$z=0+ct$$ with
$$3a+4b+2c=0$$
and
$$at-2bt+ct=0$$
thus
$$a=2b-c $$ and
$$6b-3c+4b+2c=0$$ which gives
$$c=10b $$ and
$$a=-8b $$
finally, we get
$$x=-8t $$
$$y=t $$
$$z=10t $$
A: Since the line, which I shall dub $\mathbf l$, is contained in the plane $P$,
$\mathbf l \subset P = \{(x, y, z) \in \Bbb R^3 \mid x - 2y + z = 0 \}, \tag 1$
it is orthogonal to the vector
$\mathbf n = (1, -2, 1), \tag 2$
which itself is normal to $P$; indeed, it is evident that the "vectors", that is, the points $(x, y, z)$, of $P$ satisfy
$\mathbf n \cdot (x, y, z) = (1, -2, 1) \cdot (x, y, z) = x - 2y + z = 0; \tag 3$
thus, since any tangent vector $\mathbf t \ne 0$ to $\mathbf l$ may also be thought of as a vector or point in $P$,
$\mathbf t \cdot \mathbf n = 0. \tag 4$
We are also given that the vector $\mathbf v = (3, 4, 2)$ is normal to $\mathbf l$, and hence to $\mathbf t$:
$\mathbf t \cdot \mathbf v = 0; \tag 5$
we note that $\mathbf n$ and $\mathbf v$ are not co-linear; that is, there exists no $c \in \Bbb R$ such that
$\mathbf v = c \mathbf n; \tag 6$
this is equivalent to the linear independence of $\mathbf v$ and $\mathbf n$, as is well-known; therefore the vector $\mathbf v \times \mathbf n \ne 0$, and since
$\mathbf v \cdot \mathbf v \times \mathbf n = \mathbf n \cdot \mathbf v \times \mathbf n = 0, \tag 7$
we may take
$\mathbf t = \mathbf v \times \mathbf n = (3, 4, 2) \times (1, -2, 1) = (8, -1, -10); \tag 8$
we then see that we may write $\mathbf l$ in the parametrized form
$\mathbf l(s) = s \mathbf t + \mathbf l(0); \tag 9$
to ensure $\mathbf l(s)$ passes through $\mathbf 0 = (0, 0, 0)$ there must be some $s_0$ with
$\mathbf l(s_0) = s_0 \mathbf t + \mathbf l(0) = \mathbf 0; \tag{10}$
we may take $s_0 = 0$ and $\mathbf l(0) = \mathbf 0$; then
$\mathbf l(s) = s \mathbf t \tag{11}$
satisfies all the required criteria, as is clear from the above; indeed
$\mathbf l(s) \cdot \mathbf v = s \mathbf t \cdot\mathbf v = 0,  \tag{12}$
$\mathbf l(s) \cdot \mathbf n = s \mathbf t \cdot \mathbf n = 0, \tag{13}$
and
$\mathbf l(0) = \mathbf 0. \tag{14}$
