<s>[INST] Intersection between two planes and a line? What is the coordinates of the point where the planes: $3x-2y+z-6=0$ and $x+y-2z-8=0$ and the line: $(x, y, z) = (1, 1, -1) + t(5, 1, -1)$ intersects with eachother?
I've tried letting the line where the two planes intersect eachother be equal to the given line, this results in no solutions.
I have tried inserting the lines x, y and z values into the planes equations, this too, results in no solutions.
According to the answer sheet the correct solution is: $\frac{1}{2}(7,3,-3)$
 [/INST] Isolate a variable in the planar equations and set the resulting expressions equal to each other (because they intersect):
$z= 6+2y-3x$, and $z=\frac{x+y-8}{2}$
so $6+2y-3x=\frac{x+y-8}{2}$.
Solving for x yields:
$\frac{20+3y}{7}=x$
and setting $y$ as the parameter ($y=s$), and substituting back into the original equation we have the equation of the line of intersection:
$x = \frac{20+3s}{7}$
$y=s$
$z=6+2s-\frac{60+9s}{7}$
Now, if the lines intersect the formulas for x and y must be equal so:
$s=1+t$
$\frac{20+3s}{7}=1+5t$
Solving yields $s=1.5$, which when plugged into our formulas for x,y, and z in terms of s yields $(3.5,1.5,-1.5)$, as desired.
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