The distance of the origin from the plane passing through the point `(-1,-2,-1)` and whose normal is parallel to the line ,`(x-alpha)/(1)=(y-beta)/(7)=(z-gamma)/(-5)` is

The distance of the point (1, 3, -7) from the plane passing through the point (1, -1, -1) having normal perpendicular to both the lines (x-1)/(1)=(y+2)/(-2)=(z-4)/(3) and (x-2)/(2)=(y+1)/(-1)=(z+7)/(-1) is

find the equation of a plane passing through the points (2,-1,0) and (3,-4,5) and parallel to the line 2x=3y=4z

Find the equation of the plane passing through the point (1,-2,7) and parallel to the plane 5x+4y-11z=6 .

Find the equation of the plane through the line (x-x_1)/l_1=(y-y_1)/m_1=(z-z_1)/n_1 and parallel to the line (x-alpha)/l_2=(y-beta)/m_2=(z-gamma)/n_2

The distance of the point (1,-2,4) from the plane passing through the point (1,2,2) perpendicular to the planes x-y+2z=3 and 2x-2y+z+12=0 is

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

Find the equation of a plane passing through the point (2,-1,5), perpendicular to the plane x+2y-3z=7 and parallel to the linne (x+5)/3=(y+1)/-1=(z-2)/1 .

The equation of the plane parallel to the plane 2x+3y+4z+5=0 and passing through the point (1,1,1) is

The distance of the point (1,2,3) from the plane x+y+z=11 measured parallel to the line (x+1)/(1)=(y-12)/(-2)=(z-7)/(2) is