A) \[\frac{1}{2}\]
B) \[\frac{3}{2}\]
C) \[3\]
D) \[4\]
E) \[\frac{5}{2}\]
Correct Answer: B
Solution :
Comparing the given equation of plane \[3x+y+2z+6=0\]with \[lx+my+nz+d=0\] \[\Rightarrow \] \[l=3,m=1,n=2\] Also, comparing given equation of line \[\frac{x-\frac{1}{3}}{\frac{2b}{3}}=3-y=\frac{z-1}{a}\] with\[\frac{x-{{x}_{1}}}{{{a}_{1}}}=\frac{y-{{y}_{1}}}{{{b}_{1}}}=\frac{z-{{z}_{1}}}{{{c}_{1}}},\]we get \[{{a}_{1}}=\frac{2b}{3},{{b}_{1}}=-1,{{c}_{1}}=a\] For parallel line \[l{{a}_{1}}+m{{b}_{1}}+n{{c}_{1}}=0\] \[\Rightarrow \] \[3.\frac{2b}{3}+1.(-1)+2a=0\] \[\Rightarrow \] \[2a+2b=1\] \[\Rightarrow \] \[a+b=\frac{1}{2}\] \[\Rightarrow \] \[3a+3b=\frac{3}{2}\]
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