KEAM 2011 Math Paper

The equation of plane passing through (1,0,0) is
since, plane also passing through (0,2,0).
⇒‌‌a=2b....(ii)
Given, distance from origin to plane (i) =

6

7

⇒‌‌|‌

−a+0+0

√a2+b2+c2

|=‌

6

7

⇒‌‌‌

a

√a2+b2+c2

=‌

6

7

⇒‌‌‌

2b

√4b2+b2+c2

=‌

6

7

  [from Eq.(ii)]
⇒‌‌14b=6√5b2+c2
Squaring on both sides;
⇒‌‌196b2=36(5b2+c2)
⇒‌‌196b2=180b2+36c2
⇒‌‌16b2=36c2
⇒‌‌4b=6c....(iii)
From Eq. (ii) and Eq. (iii),
2a=4b=6c
⇒‌‌‌

a

6

=‌

b

3

=‌

c

2

So, required equation of plane is,
6(x−1)+3(y−0)+2(z−0)=0
or ‌‌6x+3y+2z−6=0