A) \[2x-y-10z=9\]
B) \[2x-y+7z=11\]
C) \[2x-y+10z=11\]
D) \[2x-y-9z=10\]
Correct Answer: C
Solution :
Equation of a plane through the line of intersection of the planes \[x+2y=3,y-2z+1=0\]is \[(x+2y-3)+\lambda (y-2z+1)=0\] \[\Rightarrow \]\[x+(2+\lambda )y-2\lambda (z)-3+\lambda =0\] (i) Now, plane (i) is \[\bot \] to x + 2y = 3 \[\therefore \]Their dot product is zero i,e. \[1+2(2+\lambda )=0\Rightarrow \lambda =-\frac{5}{2}\] Thus, required plane is \[x+\left( 2-\frac{5}{2} \right)y-2\times \frac{-5}{2}(z)-3-\frac{5}{2}=0\] \[\Rightarrow \]\[x-\frac{y}{2}+5z-\frac{11}{2}=0\]\[\Rightarrow \]\[2x-y+10z-11=0\]
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