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=`

What is the distance of the point (-1,2,1) from the plane 2x-3y+4z+5=0.

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

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

If the distance of the point (1,1,1) from the origin is half its distance from the plane x+y+z+k=0 , then k is equal to

If the distance of the point (1,1,1) from the origin is half its distance from the plane x+y+z+k=0 , then k is equal to

If the distance of the point ( 1,1,1) from the origin is half its distance from the plane x+y+z+k=0 , then k is equal to :

If the distance of the point (1,1,1) from the origin is half of its distance from the plane x + y + z + k = 0, then the value of k are

Locus of the image of the point (1,1) in the line (5x-2y-7)+lambda(2x-3y+6)=0lambda in R, is

If the distance of the plane x - y + z + lambda = 0 from the point (1, 1, 1) is d_1 and the distance of this point from the origin is d_2 and d_2d_2 = 5 then find the value of lambda .