MHT CET 2020 Math Solved 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−3

−6

=r (say) . . . (i)
Then, any point on Eq. (i) 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)=5
⇒−7r+6=5⇒r=‌

1

7

Since, the point is (‌

9

7

,−‌

11

7

,‌

15

7

).
Distance between (1,−2,3) and (‌

9

7

,−‌

11

7

,‌

15

7

)
=√(‌

4

49

+‌

9

49

+‌

36

49

)=√(‌

49

49

)=1