The equation of a plane is `2x + y +z = 0` and the coordinates of point P are `(3,5,7)`. The

Find the equation of the plane passing through the intersection of the planes 2x + y + 3z-7=0 and 2x + 5y + 3z - 9 = 0 and the point (2,1,3).

Find the equation passing throught the intersection of the planes : 2x-7y+4z=3 and 3x-5y+4z+11=0 and the point (-2,1,3).

Find the image of the point P(3,5,7) in the plane 2x+y+z=0

Find the value of p so that the lines (x+1)/(-3)=(y-p)/(2)-(z+2)/(1) and (x)/(1)=(y-7)/(1)=(z+7)/(2) are in the same plane.for this value of p,find the coordinates of their point of intersection and the equation of the plane containing them.

Find the value of p so that the lines (x+1)/(-3)=(y-p)/(2)=(z+2)/(1) and (x)/(1)=(y-7)/(1)=(z+7)/(2) are in the same plane for this value of p,find the coordinates of their point of intersection and the equation of the plane containing them.

(i) Find the equation of the plane passing through the intersection of the planes : 2x - 7y + 4z = 3 and 3x - 5y + 4z + 11 = 0 and the point (-2, 1,3). (ii) Find the equation of plane through the intersection of planes : 3x - y + 2z - 4 = 0 and x + y + z - 2 = 0 and the point (2,2,1) .

If the equation of a plane P , passing through in the intersection of the planes, x+4y-z+7=0 and 3x+y+5z=8 is ax+by+6z=15 for some a,b,c in R , then the distance of the point (3,2,-1) form the plane P is ............