Finding 3 lines on a plane such that the area of the triangle is $\sqrt{6}$ Given $ H=\{(x,y,z) : -x+2y-z=2\} $
I'm trying to find the three sides of a triangle on H with the area of  $\sqrt{6}$
I guess I only need 2 perpendicular vectors on that plane with magnitudes $||v_1|| ||v_2||=2\sqrt{6}$ to define a right triangle which would be easier to manage but I'm not sure how to select them.

You are on the right track. Consider the parallel plane $H_0$ given by the equation $-x+2y-z=0$ and the vector $v_1=(1,1,1)$ which lies in that plane. Then find a vector $v_2=(2a,a+b,2b)\in H_0$ such that $v_1\perp v_2$ that is $0=v_1\cdot v_2=3a+3b$ and $\|v_2\|=\frac{2\sqrt{6}}{\|v_1\|}=2\sqrt{2}$. The required triangle is given by $A=(0,1,0)$, $B=A+v_1$ and $C=A+v_2$.
