Question

The square of the distance of the point of intersection of the line \(\cfrac{x-1}{2}\)= \(\cfrac{y-2}{3}\)= \(\cfrac{z+1}{6}\)and the plane 2x – y + z = 6 from the point (–1, –1, 2) is _____.

Answer 1

  \(\cfrac{x-1}{2}\)= \(\cfrac{y-2}{3}\)= \(\cfrac{z+1}{6}\) = \(\lambda\)