`4-(2(z-4))/3=1/2(2z+5)`

find z . 5-2((z-4)/3) =1/2(2z +3)

" The point of intersection of lines "(x-1)/(2)=(y-2)/(3)=(z-3)/(4)" and "(x-4)/(5)=(y-1)/(2)=(z)/(1)" is "

Which of the following line is passing through point of intersection of line (x-4)/(2)=(y-3)/(2)=(z-5)/(1) and the plane x+y+z=2( a) (x)/(2)=(y+1)/(3)=(z-3)/(4) (b) (x-2)/(3)=(y-1)/(3)=(z-2)/(2)(c)(x-2)/(2)=(y+1)/(3)=(z-2)/(4)(d)(x-2)/(2)=(y+1)/(3)=(z+3)/(4)

Show that the lines (x-1)/(2)=(y-2)/(y-1)=(z-3)/(4) and (x-4)/(5)=(y-1)/(2)=z intersect

Show that the lines : (i) (x -5)/(7) = (y + 2)/(-5) = (z)/(1) " " and (x)/(1) = (y)/(2) = (z)/(3) (ii) (x - 3)/(2) = (y + 1)/(-3) = (z - 2)/(4) and (x + 2)/(2) = (y - 4)/(4) = (z + 5)/(2) are perpendicular to each other .

If z_1,z_2,z_3,z_4 be the vertices of a parallelogram taken in anticlockwise direction and |z_1-z_2|=|z_1-z_4|, then sum_(r=1)^4(-1)^r z_r=0 (b) z_1+z_2-z_3-z_4=0 a r g(z_4-z_2)/(z_3-z_1)=pi/2 (d) None of these

If (5z_(1))/(7z_(2)) is purely imaginary then |(2z_(1)+3z_(2))/(2z_(1)-3z_(2))|=