The perpendicular distance of the plance ax + b y + c z + d = 0 from the point `P(x_(0) , y_(0) , z_(0))` is `(ax_(0)+by_(0)+cz_(0)+d)/sqrt(a^(2)+b^(2)+c^(2))`

The perpendicular distance from the point (-2,3, 1) to the plane 2x – 3y – 6z + 5 = 0 is

The plane x =0, x =a, y =0, y =a , z = and z =a from a

If Q(h,k) is the foot of the perpendicular of P(x_(1),y_(1)) on the line ax+by+c=0 then prove that (h-x_(1)),a=(k-y_(1)),b=-(ax_(1)+by_(1)+c):(a^(2)+b^(2)) .

If the distance between the planes 4x -2y - 4z +1 =0 and 4x -2y -4z + d=0 is 7 , then d is

If a , b , c in R such that a + b + c= 0 and az_(1) + bz_(2) + cz_(3) = 0 then z_(1) , z_(2) , z_(3) are

Assertion (A) : The normals to the planes x-lambday+z=1, 3x+2y-lambdaz+1=0 are perpendicular to each other if lambda+1=0 . Reason (R ) : If the planes a_(1)x+b_(1)y+c_(1)z+d_(1)=0,a_(2)x+b_(2)y+c_(2)z+d_(2)=0 are perpendicular to each other then a_(1)a_(2)+b_(1)b_(2)+c_(1)c_(2)=0