Answer:
Option A
Explanation:
Equation of plane passing through (2,2,1) is
a(x-2) +b(y-2)+c(z-1)=0 .....(i)
Since , above plane is perpendicular to
3x+2y+4z+1=0
and 2x+y+3z+2=0
$\therefore$ 3a+2b+4c=0 ......(ii)
and 2a+b+3c=0 .....(iii)
[ $\because$ for perpendicular , a1a2+b1b2+c1c2=0]
On multiplying eq.(iii) by 2 , we get
4a+2b+6c=0 ......(iv)
On subtracting eq.(iv) from eq.(ii) , we get
$\Rightarrow c=\frac{-a}{2}$
On putting $ c=\frac{-a}{2}$ in eq. (iii) , we get b = $\frac{-a}{2}$
on putting b = $\frac{-a}{2}$ and $c=\frac{-a}{2}$ in eq.(i)
we get $a(x-2)-\frac{a}{2}(y-2)-\frac{a}{2}(z-1)=0$
$\Rightarrow\frac{a}{2}[2(x-2)-(y-2)-(z-1)]=0$
$\Rightarrow2x-4+y+2-z+1=0$
$\Rightarrow2x-y-z-1=0$
