JEE Mains Model Paper 15

Any point P on the given line is
(r,2r + 1,3r + 2) direction ratios of AP are
(r - 1, 2r - 5 ,3r - 1)
Now AP ⊥ to the given line if
1(r - 1) + 2( 2r - 5) + 3 (3r - 1) = 0
⇒ 14r -14 = 0
⇒ r = 1 ∴ P is (1,3,5).
Thus the foot of the ⊥ from A on the line is (1,3,5). Let be the image of A in the given
line. Then P is the mid point of AB.
∴

a+1

2

= 1 ⇒ a = 1 and

c+3

2

= 5 ⇒ c = 7

b+6

2

= 3 ⇒ b = 0
∴ B is (1, 0, 7)