How to find the vector equation of a plane given the scalar equation? [closed]

How would I find the vector equation of the plane: $x + 2y + 7z - 3 = 0$

So far, I found the normal vector: it's $(1, 2, 7)$.

The vector equation of a plane has the form $\mathbf r(s,t) = \mathbf r_0 + s\mathbf u + t\mathbf v$.

To find $\mathbf u$ and $\mathbf v$, you need two vectors (which are not collinear) which are orthogonal to $(1,2,7)$ (do you see why they need to be orthogonal to this vector?).

So let's find one by simply plugging some values into the definition of orthogonality: $(a,b,c)\cdot(1,2,7)=0$. Find just $1$ nonzero solution to that and you have your $\mathbf u$. Then take the cross product of $\mathbf u$ and $(1,2,7)$ to get your $\mathbf v$.

Then we need to get your $\mathbf r_0$. To do so, just plug in two arbitrary values into your equation $x + 2y + 7z - 3 = 0$ and solve for the third. Then you'll have your $\mathbf r_0$ and you're done. :)

Hint: To find directions, you can consider $x+2y+7z = 0$. The number $3$ is just a translation upwards. Write: $$z = \frac{-x}{7}-\frac{2y}{7} \implies (x,y,z) = \left(x,y, \frac{-x}{7}-\frac{2y}{7}\right) = x\left(1,0,-\frac{1}{7}\right) + y\left(0,1,-\frac{2}{7}\right).$$