BITSAT 2005 Solved Question Paper

The equation of any plane through (2,−1,3) is
a(x−2)+b(y+1)+c(z−3)=0 ........(i)
where a,b and c are direction ratios, Since Eq. (i) is parallel to

→

a

and

→

b

∴ 3a+0b−c=0 ........(ii)
and −3a+2b−2c=0 .......(iii)
Solving Eqs. (ii) and (iii), we get

a

2

=−

b

6−3

=

c

6

=k (say)
⇒a=2k,b=−3k,c=6k
Putting the values of a,b and c in Eq. (i), we get
2k(x−2)−3k(y+1)+6k(z−3)=0
⇒2x−3y+6z−25=0
which is a required equation of a plane.