Find intersection of planes given by $x+y+z+1=0$ and $x + 2y + 3z +4=0$.

**Solution**: In three dimensions (which we are implicitly working
with here), what is the intersection of two planes? As long as the
planes are not parallel, they should intersect in a line. So our
result should be a line.

How does one write an equation for a line in three dimensions? You should convince yourself that a graph of a single equation cannot be a line in three dimensions. Instead, to describe a line, you need to find a parametrization of the line.

How can we obtain a parametrization for the line formed by the intersection of these two planes? We need an equation for all $(x,y,z)$ that satisfy both $x+y+z+1=0$ and $x + 2y + 3z +4=0$. We start be attemping to solve this system of two equations. Since this is a system of two equations and three unknowns, we know we can't solve it for a unique $(x,y,z)$. But this is consistent with our above conclusion that the intersection is a line, not a point. We want a whole bunch of $(x,y,z)$ that satisfy the two equations.

We could use either the method of “substitution” or the method of “elimination” to solve these equations. We'll use the method of elimination. Since in both equations, $x$ has a coefficient of 1, we'll subtract the two equations to eliminate $x$. If we subtract $x+y+z+1=0$ from $x + 2y + 3z +4=0$, we are left with \begin{align*} y+2z+3=0 \end{align*} which we can rewrite as $y=-2z-3$.

We don't have enough equations to uniquely solve for either $y$ or $z$. We'll simply set $z=t$, where $t$ can be any real number. The variable $t$ will be our free parameter for the parametrization of the line. Since we found that $y=2z-3$, we can write it in terms of $t$ as $y=-2t-3$. To find $x$ in terms of $t$, we plug these values of $y$ and $z$ into one of the original equations. We'll use the first one so that \begin{align*} x=-y-z-1=2t+3-t-1=t+2. \end{align*} We can write our parametrization of the line as \begin{align*} (x,y,z) = (t+2, -2t-3, t). \end{align*}