Question

If the plane $ 3x+y+2z+6=0 $ is parallel to the line $ \frac{3x-1}{2b}=3-y=\frac{z-1}{a}, $ then the value of $ 3a+3b $ is

• A. $ \frac{1}{2} $
• B. $ \frac{3}{2} $
• C. $ 3 $
• D. $ 4 $

Answer 1

The Correct Option is B

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} $