Question

The plane passing through the point $(-2,-2-3)$ and containing the line through the points $(1,1,1)$ and $(1,-1,2)$ make intercepts $a,b,c$ along the $x,y,z$ axis respectively. Then the value of $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$ is:

• A. $3$
• B. $4$
• C. $6$
• D. $1$

Answer 1

The correct option is D $1$

Let the given points be $A(-2,-2-3)$, $B(1,1,1)$ and $C(1,-1,2)$ so that $AB$ and $AC$ are two lines on the required plane.
DR's of $AB$ and $AC$ are $(3,3,4)$ and $(3,1,5)$ respectively.
Hence equation of the plane is:
$\left|\begin{array}{ccc}x+2& y+2& z+3\\ 3& 3& 4\\ 3& 1& 5\end{array}\right|=0$
$\Rightarrow 11(x+2)-3(y+2)-6(z+3)=0$
$\Rightarrow 11x-3y-6z=2$
Hence, we have $a=\frac{2}{11}.\frac{-2}{3},\frac{-2}{6}$
So, $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{11}{2}+\frac{-3}{2}+\frac{-6}{2}=\frac{2}{2}=1$
$\therefore \frac{1}{a}+\frac{1}{b}+\frac{1}{c}=1$