JEE Main 24 Feb 2021 Shift 1 Solved Paper

Given, P(1,1,9). Equation of plane
x+y+z‌‌=17
‌ Equation of line ‌⇒‌

x−3

1

‌‌=‌

y−4

2

=‌

z−5

2

⇒‌‌‌

x−3

1

=‌

y−4

2

=‌

z−5

2

=λ (let)
⇒‌‌x=λ+3;y=2λ+4;z=2λ+5
∴ The point we have is (λ+3,2λ+4,2λ+5).
∵ This point lies on the plane
x+y+z=17.
∴‌‌λ+3+2λ+4+2λ+5=17
⇒‌‌‌‌λ=1
∴ The coordinate of point is (4,6,7)
∴ Required distance between (1,1,9) and (4,6,7) is
‌‌=√(4−1)2+(6−1)2+(7−9)2
‌‌=√9+25+4=√38