VITEEE 2017 Paper

Equation of the line through (1, -2, 3) parallel to the line

x

2

=

y

3

=

z−1

−6

is

x−1

2

=

y+2

3

=

z−1

−6

= r (say) ... (1)
Then any point on (1) is (2r + 1,3r -2, -6r + 3)
If this point lies on the plane x - y + z = 5 then
(2r + 1) - (3r - 2) + (-6r + 3) = 2 ⇒ r =

1

7

Hence the point is
(97,

−11

7

,

15

7

)
Distance between (1, -2, 3) and (97,

−11

7

,

15

7

)
= √(

4

49

+

9

49

+

36

49

) = √(

49

49

) = 1