Find an equation of the plane that passes through the point $(1, 3, 1)$ and contains the line: $x = t, y = t$ and $z = -2 + t$.
My attempt:
The points are $(1,3,1), (0,0,-2)$ and if $t = 1$: $(1,,1,-1)$
$a = (1,1,-1) - (0,0,-2) = [1,1,1]$
$b = (1,1,-1) - (1,3,1) = [0,-2,-2]$
normal vector is $n = a \times b = 2j - 2k = [0,2,-2]$
Therefore the equation is found by $[0,2,-2] \cdot [x-1, y-3, z-1] = 0$
$\leftrightarrow 2y -2z = 4$
$\leftrightarrow y - z = 2$
right?