KCET 2020 Math Solved Paper

Let

x−3

2

=

y−3

3

=

z+5

6

=t
⇒(x,y,z)=(2t+3,3t+3,6t−5)
∴ d.r.'s of the line perpendicular to

x−3

2

=

y−3

3

=

z+5

6

and joining (2𝑡 + 3, 3𝑡 + 3, 6𝑡 − 5)
and (1, 2, −4) is (2𝑡 + 2, 3𝑡 + 1, 6𝑡 − 1)
∴ 2(2𝑡 + 2) + 3(3𝑡 + 1) + 6(6𝑡 − 1) = 0
⇒t=−

1

49

∴ Distance =√(2t+2)2+(3t+1)2+(6t−1)2
=√49t2+2t+6
=√

1

49

−

2

49

+6=

√293

7