JEE Main 26 Feb 2021 Shift 2 Solved Paper

Given, x+2y+z=6 . . . (i)
and ‌‌y+2z=4 . . . (ii)
Put y=4−2z from Eq. (ii) Eq. in (i), we get
‌x+8−4z+z‌‌=6
⇒‌x‌‌=−2+3z
⇒‌‌

x+2

3

‌‌=z . . . (iii)
‌y‌‌=4−2z
⇒‌‌

y−4

−2

‌‌=z . . . (iv)
From Eqs. (iii) and (iv), line of intersection of two planes is
‌

x+2

3

=‌

y−4

−2

=‌

z

1

=λ
Then, Direction ratios of PQ is
x−3λ−4−2λ,z=λ
(3λ−5,−2λ+2,λ−1)
(3λ−5,−2λ+2,λ−1)Since, PQ is perpendicular to the line, then
3(3λ−5)−2(−2λ+2)+1(λ−1)=0
∴‌‌λ=‌

10

7

‌
∴‌‌P(‌

16

7

,‌

8

7

,‌

10

7

)‌
‌ Then, ‌21(α+β+γ)‌‌=21(‌

16

7

+‌

8

7

+‌

10

7

)
‌‌=21(‌

34

7

)=3×34=102