Answer
The intersection of the planes $x+y+z=1$ and $3x-2y+z=5$ is the line parametrized by
$\left( {x,y,z} \right) = \left( {t,\frac{1}{3}\left( {2t - 4} \right),\frac{1}{3}\left( { - 5t + 7} \right)} \right)$
Work Step by Step
We have two equations of the planes: $x+y+z=1$ and $3x-2y+z=5$.
Step 1. Find $y$ in terms of $x$.
From the first equation we get $z=1-x-y$. Substituting it in the second equation gives
$3x-2y+z=5$
$3x-2y+1-x-y=5$
$2x-3y=4$
So, $y = \frac{1}{3}\left( {2x - 4} \right)$.
Step 2. Find $z$ in terms of $x$.
From the first equation we get $y=1-x-z$. Substituting it in the second equation gives
$3x-2y+z=5$
$3x-2(1-x-z)+z=5$
$5x+3z=7$
So, $z = \frac{1}{3}\left( { - 5x + 7} \right)$.
Step 3. Write $x=t$, so $y = \frac{1}{3}\left( {2t - 4} \right)$ and $z = \frac{1}{3}\left( { - 5t + 7} \right)$.
The intersection of the planes $x+y+z=1$ and $3x-2y+z=5$ is the line parametrized by
$\left( {x,y,z} \right) = \left( {t,\frac{1}{3}\left( {2t - 4} \right),\frac{1}{3}\left( { - 5t + 7} \right)} \right)$
