The problem is to find the distance between the point $(3, -2, 4)$ and the plane $2x-5y+z=10$ I tought of doing it like this:
The shortest line between the point and the plane is orthogonal to the plane, so it's direction is $<2, -5, 1>$, and a point on this line is $(3, -2, 4)$ . That means that the line has the parametric equations $x=3+2t, y=-2-5t, z=4+t$. I plugged in $x, y, z$ into the plane equation and got $\dfrac{1}{11}$ for $t$. This means that the distance between the plane and the point should be the distance between the point $(3, -2, 4)$ and the point $x=3+2(\dfrac{1}{11}), y=-2-5(\dfrac{1}{11}), z=4+(\dfrac{1}{11})$
I got $\dfrac{\sqrt{29}}{11}$ as the distance, but WolframAlpha says its $\sqrt{\frac{10}{3}}$
I get $t=-\frac{1}{3}$, not $\frac{1}{11}$ ...