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

Answer: (B)

$\frac{3}{2}$

Answer 2

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 $l{{a}_{1}}+m{{b}_{1}}+n{{c}_{1}}=0$
$\Rightarrow$ $3.\frac{2b}{3}+1.(-1)+2a=0$
$\Rightarrow$ $2a+2b=1$
$\Rightarrow$ $a+b=\frac{1}{2}$
$\Rightarrow$ $3a+3b=\frac{3}{2}$