A) \[2x-y-10z=9\]
B) \[2x-y-9z=10\]
C) \[2x-y+10z=11\]
D) \[2x-y+7z=11\]
Correct Answer: C
Solution :
Let the plane be \[{{P}_{1}}+\lambda {{P}_{2}}=0\] \[\Rightarrow (x+2y-3)+\lambda (y-2z+1)=0\] \[\Rightarrow x+(2+\lambda )y-2\lambda z+(\lambda -3)=0\] Now \[1(1)+2(2+\lambda )=0\] \[\Rightarrow 5+2\lambda =0\] \[\Rightarrow \lambda =\frac{-5}{2}\] \[\Rightarrow x+\left( 2-\frac{5}{2} \right)y+5z+\left( \frac{-5}{2}-3 \right)=0\] \[\Rightarrow x-\frac{y}{2}+5z-\frac{11}{2}=0\] \[\Rightarrow 2x-y+10z=11\]
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