Construct a line equidistant from two perpendicular planes

$AB=(1,-2,2)$. Thus $\mathbf r=(0,2,0)+\lambda(1,-2,2)$, or $\mathbf r=(1,0,2)+\lambda(1,-2,2)$, λ real.

$[(0,2,0)+\lambda(1,-2,2)]\cdot(2,2,1)=2\lambda+4-4\lambda+2\lambda=4$, so every point on AB lies in π1.

$(0,2,0)\cdot(2,2,1)=4$, so A lies in π1; also $(1,-2,2)\cdot(2,2,1)=2-4+2=0$. Thus the line is parallel to the plane and shares A with it, proving containment. The source repeats the label Method 1 for this second method.

Cross AB’s direction with π1’s normal:$(1,-2,2)\times(2,2,1)=(-6,3,6)=3(-2,1,2)$. Through A, π2:$\mathbf r\cdot(-2,1,2)=(0,2,0)\cdot(-2,1,2)=2$, as required.

$AP=(p,-2,0)$. Both normals have magnitude 3. Equal distances require $|AP\cdot(2,2,1)|/3=|AP\cdot(-2,1,2)|/3$, hence $|2p-4|=|-2p-2|$. One branch gives $2p-4=-2p-2$, so p=1/2. The other gives $2p-4=2p+2$, which has no solution.

Take the line through P parallel to AB:$\mathbf r=(1/2,0,0)+\mu(1,-2,2)$, μ real. Its direction is perpendicular to both normals, so each plane-distance is constant. At P both distances equal 1, nonzero; the whole line is therefore equidistant and meets neither plane.