Can there be multiple Cartesian equations of a plane?

Question

I want to find the Cartesian equation of this plane given in the parametric form:

$$ \begin{cases} \begin{align} x-1&=-s+t\\ y-2&=-s-3t\\ z&=s-3t \end{align} \end{cases} $$

Which is equivalent to

$$ \begin{cases} \begin{align} x-1&=-s+t\\ -x+y-1&=-4t\\ x+z-1&=-2t \end{align} \end{cases} $$

Now, from here, I'm tempted to multiply the second equation by $-2$ and subtract this from the third equation. This would give me:

$$ \begin{cases} \begin{align} x-1&=-s+t\\ -x+y-1&=-4t\\ 3x-2y+z+1&=0 \end{align} \end{cases} $$

However, the answer key gives me:

$$ \begin{cases} \begin{align} x-1&=-s+t\\ -x+y-1&=-4t\\ 3x-y+2z-1&=0 \end{align} \end{cases} $$

Which I believe comes from multiplying the third equation by $2$ and subtracting the second equation from that.

My question is, are both

$$ 3x-2y+z=-1 $$ and $$ 3x-y+2z=1 $$

correct as the Cartesian equation of the plane?

Answer 1

Check:

$$3(-s+t+1)-2(-s-3t+2)+(s-3t)=\color{red}{6t-1}$$

and

$$3(-s+t+1)-(-s-3t+2)+2(s-3t)=1.$$

Answer 2

You made a mistake in your computations. If you do what you wrote, you cannot possibly get $0$ after the $=$ sign. Instead, you should do:$$-(-x+y-1=-4t)+2(x+z-1=-2t)=3x-y+2z-1=0.$$

Answer 3

In your second system: $$(2)\qquad \begin{cases}x-1=-s+t\\-x+y-1=-4t\\x+z-1=-2t\end{cases}$$ performing the operation you describe does not give you the third equation in your third system: $$(3)\qquad \begin{cases}x-1=-s+t\\-x+y-1=-4t\\\color{red}{3x-2y+z+1=0}\end{cases}$$ What you could do is subtract the second equation from $2$ times the third equation to get: $$(3')\qquad \begin{cases}x-1=-s+t\\-x+y-1=-4t\\\color{blue}{3x-y+2z-1=0}\end{cases}$$