The image of the point `(-1,3,4)` in the plane `x-2y=0` is a. `(-(17)/3,-(19)/3,4)` b. `(15 ,11 ,4)` c. `(-(17)/3,-(19)/3,1)` d. `(9/5,(-13)/5,4)`

then image of the point (-1,3,4) in the plane x-2y=0

The image of the point (1, 3, 4) in the plane 2x-y+z+3=0 is (3,\ 5,\ 2) b. (-3,5,\ 2) c. (3,\ 5,\ -2) d. (3,\ -5,\ 2)

The image of (4,5,-3) in the X-Y plane is (4,5,3)

Find the image of the point (5,9,3) in the line (x-1)/(2)=(y-2)/(3)=(z-3)/(4)

The reflection of the point (2,-1,3) in the plane 3x-2y-z=9 is (A) (28/7,15/7,17/7) (B) (26/7,-15/7,17/7) (C) (15/7,26/,-17/7) (D) (26/7,17/7,-15/70)

Statement-I The point A(3, 1, 6) is the mirror image of the point B(1, 3, 4) in the plane x-y+z=5 . Statement-II The plane x-y+z=5 bisect the line segment joining A(3, 1, 6) and B(1,3, 4) .

The line passing through the points (5,1,4) and (3,b,1) crosses the yz-plane at the point (0,(17)/(2), - (13)/(2)). Then:

The line (x-3)/1=(y-4)/2=(z-5)/2 cuts the plane x+y+z=17 at

Let P=0 be the equation of a plane passing through the line of intersection of the planes 2x-y=0 and 3z-y=0 and perpendicular to the plane 4x+5y-3z=8. Then the points which lie on the plane P=0 is/are a.(0,9,17) b.(1/7,21/9) c.(1,3,-4) d.(1/2,1,1/3)

The points (3,-2,-1),(-1,1,2),(2,3,-4) and (4,5,lamda) are coplanar when lamda= (A) 0 (B) (-146)/17 (C) 1 (D) (-17)/9