Find the shortest distance from the point (2, 10, 1) to the plane `vecr ( 3 hati - hatj + 4hatk ) = 2sqrt(26)`

Find the distance of the point (2, 1,-1) from the plane vecr.(hati-2hatj+4hatk)=9 .

Find the perpendicular distance of the point (2, -3, 3) from the plane vecr*(hati-2hatj-2hatk)=8 .

Find the distance of the point (-1,-2,3) from the plane vecr.(2hati-3hatj+4hatk)=4

Find the distance of a point (2,5,-3) from the plane vecr.(6hati-3hatj+2hatk)=5

The vector equation of the plane through the point (2,1,-1) and parallel to the plane vecr.(hati+3hatj-hatk)=0 is

Find the direction cosines of the unit vector perpendicular to the plane. vecr(6hati-3hatj-2hatk)+1=0

Find the direction cosines of the normal to the plane and length of the perpendicular from the origin to the plane vecr.(3hati-4hatj+12hatk)=5 .

Find the coordinates of the foot of the perpendicular from the point (5, 2, -8) to the straight line vecr=2hati-hatj+2hatk+t(hati+3hatj-hatk) . Find also the shortest distance from the point to this straight line.