If the distance of the point (1,1,1) from the plane `x-y+z+ lamda =0 " is " (5)/(sqrt3)` find the value (s) of `lamda`

The distance of the point (2,1,0) from the plane 2x+y+2z+5=0

Find the distance of the point (1,2,0) from the plane 4x+3y+12z+16=0

Find the distance of the point (1, 2-1) from the plane x -2y + 4z - 10 = 0 .

If the product of distances of the point (1,1,1) from the origin and plane x-y+z+lambda=0 be 5 then lambda=

A plane P = 0 passing through the point (1, 1, 1) is perpendicular to the planes 2x-y+2z=5 and 3x+6y-2z=7 . If the distance of the point (1, 2, 3) from the plane P = 0 is k units, then the value of 34k^(2) is equal to

The plane x-y-z=2 is rotated through 90^(@) about its line of intersection with the plane x+2y+z=2 . The distance of (-1,-2,-1) from the plane in the new position is lamda sqrt(6/7) . Then the value of lamda is equal to

Find the perpendicular distance from the point (1,-3,4) to the plane 3x-4y+12z - 1 = 0 .

Find the distance of the point (1,-2,3) from the plane x-y+z=5 measured parallel to the line (x-1)/2=(y-3)/3=(z+2)/-6 .

Find the distance of the point (0,-3,2) from the plane 3x+2y+2z+5=0 , measured parallel to the line (x+1)/3=(y+1)/2=z/3 .