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Soft question:

Let $abc\neq 0$ and $$a(x-x_0)+b(y-y_0)+c(z-z_0)=0$$ Besides $$x=x_0$$ $$y=y_0$$ $$z=z_0$$ Are there any other solutions?

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    What do you mean. The whole point of such equation is any $(x,y,z)$ on the plane is (made) a solution. The solutions are points on the plane with normal vector $\langle a,b,c \rangle$ and where $(x_0,y_0,z_0)$ is a point on the plane. – 2017-02-21
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    Le me get this straight: If $2(x-3)+5(y-11)+7(z-13)=0$ then any $x,y,z$ is solution? – 2017-02-21
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    Any $(x,y,z)$ on the plane $P$ with normal vector $\langle 2,5,7 \rangle$ and a point $(3,11,13)$ lying on the plane is a solution. But of course not any $(x,y,z)$ is general. – 2017-02-21
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    Take the equation $1(x-1)+1(y-1)+1(z-1)=0$ for example and maybe take $z=0$. So now the problem becomes how many solutions to $x+y=3$ are there. You know the answer. – 2017-02-22

4 Answers 4

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There will be a whole 2 dimensional subspace of solutions. \begin{eqnarray*} x &=& x_0 + \frac{\lambda}{a} \\ y &=& y_0 + \frac{\mu}{b} \\ z &=& z_0 - \frac{\lambda+\mu}{c} \end{eqnarray*} The equation represents a plane.

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Yes! There's a whole plane of solutions, quite literally. The equation you've given can be written as ${\bf r} \cdot {\bf n} = {\bf p} \cdot {\bf n}$, where ${\bf r} = (x, y, z)$, ${\bf n} = (a, b, c)$ and ${\bf p} = (x_0, y_0, z_0)$, which is exactly the equation of a plane. (See here and here)

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Clearly so. Specify any two of $x,y,z$ arbitrarily, and then solve the simple linear equation explicitly for the third (since none of $a,b,c$ can be $0$).

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Let's perturb our proposed solution, so that $z$ changes from $z_0$ to $z_0 + \delta_z$.

Then in order to compensate the change, we need to add to $x$ and $y$ the terms $\delta_x$ and $\delta_y$, such that $a\delta_x$ + $b\delta_y$ = $c\delta_z$. To parametrize this space, we need parameters $\alpha$ and $\beta$, both real.

Therefore the solution space is $x = x_0 + \frac{\alpha}{a}$, $y = y_0 + \frac{\beta}{b}$, and $z = z_0 - \frac{\alpha + \beta}{c}$ by inspection. This is a plane, of course.