A) \[x+5y-3z=0\]
B) \[x-5y+3z=0\]
C) \[x-5y-3z=0\]
D) \[3x-10y+5z=0\]
E) \[x+5y+3z=0\]
Correct Answer: B
Solution :
The equation of the plane through given line is \[A(x-1)+B(y-2)+C(z-3)=0\] ...(i) where A, B and C are the DRs of the normal to the plane. Since, the straight line lie on the plane. .. DRs of the plane is perpendicular to the line, ie, \[5A+4B+5C=0\] ...(ii) Since, it passes through (0, 0, 0), we get \[-A-2B-3C=0\] \[\Rightarrow \] \[A+2B+3C=0\] ...(iii) On solving Eqs. (ii) and (iii), we get \[\frac{A}{2}=\frac{B}{-10}=\frac{C}{6}\] From Eq. (i), \[2(x-1)-10(y-2)+6(z-3)=0\] \[\Rightarrow \] \[2x-2-10y+20+6z-18=0\] \[\Rightarrow \] \[2x-10y+6z=0\] \[\Rightarrow \] \[x-5y+3z=0\]
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