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.