How do you find the parametric equation through the point (6, 0, 10) and parallel...

Step 1
Any plane parallel to the plane ${\displaystyle -3x+2y-4z=0}$ will have an equation of the same form but equal to a different constant:
${\displaystyle -3x+2y-4z=C\text{[1]}}$
To force the point ${\displaystyle (6,0,10)}$ to be in the plane, substitute the point into equation [1] and solve for C:
${\displaystyle -3\left(6\right)+2\left(0\right)-4\left(10\right)=C}$
${\displaystyle C=-58}$
The scalar equation of the plane parallel to the given plane is:
${\displaystyle -3x+2y-4z=-58\text{[2]}}$
The vector equation of the plane is
${\displaystyle (x,y,z)=(6,0,10)+s\left(\overline{u}\right)+t\left(\overline{v}\right)}$
Where ${\displaystyle \overline{u}\text{and}\overline{v}}$ are any two vectors in the plane.
Let's use equation [2] to find two other points in the plane:
Let ${\displaystyle x=6\text{and}y=2}$
${\displaystyle -3\left(6\right)+2\left(2\right)-4z=-58}$
${\displaystyle -4z=-44}$
${\displaystyle z=11}$
The vector from the point ${\displaystyle (6,0,10)}$ to the point ${\displaystyle (6,2,11)}$ is:
${\displaystyle \overline{u}=2\hat{j}+\hat{k}}$
Let ${\displaystyle x=10\text{and}y=0}$
${\displaystyle -3\left(10\right)+2\left(0\right)-4z=-58}$
${\displaystyle -4z=-28}$
${\displaystyle z=7}$
The vector from point ${\displaystyle (6,0,10)}$ to the point ${\displaystyle (6,0,7)}$ is:
${\displaystyle \overline{v}=4\hat{i}-3\hat{k}}$
A vector equation of the plane is:
${\displaystyle (x,y,z)=(6,0,10)+s(2\hat{j}+\hat{k})+t(4\hat{i}-3\hat{k})}$
From the above equation, we can write a set of parametric equations:
${\displaystyle x=4t+6}$
${\displaystyle y=2s}$
${\displaystyle z=s-3t+10}$
NOTE: These are not unique equations, because we could have chosen other points and obtained different vectors.