Question

Equation of the plane perpendicular to the line $\frac{x}{1}=\frac{y}{2}=\frac{z}{3}$ and Passing through the point $(2,3,4)$ is

• A. $x+2y+3z=9$
• B. $x+2y+3z=20$
• C. $2x+3y+z=17$
• D. $3x+2y+z=16$

Answer 1

Answer: (B)

Equation of plane passing through $(2,3,4)$ is
$a(x-2)+b(y-3)+c(z-4)=0$...(i)
Since, above plane is perpendicular to the line.
$\frac{x}{1}=\frac{y}{2}=\frac{z}{3}$
Thus, normal to the plane is parallel to the line.
So, DR's of normal is $(1,2,3)$, i.e., $(a,b,c)=(1,2,3)$.
Now, from Eq. (i),
$1(x-2)+2(y-3)+3(z-4)=0$
$\Rightarrow x+2y+3z=20$