Q85 / 180 · COMEDK 2014 · Mathematics
Determine the plane through the intersection of the planes $x+2 y+3 z-4=0$ and
$2 x+y-z+5=0$ and perpendicular to the plane
$5 x+3 y+6 z+8=0$
- A$-51 x-15 y-50 z-173=0$
- Correct answerB$51 x+15 y-50 z+173=0$
- C$51 x-15 y+50 z-173=0$
- D$51 x+50 y+15 z+173=0$
Solution
Equation of plane through the intersection of planes $x+2 y+3 z-4=0$ and $2 x+y-z+5=0$ is
$(x+2 y+3 z-4)+k(2 x+y-z+5)=0$
or $(1+2 k) x+(2+k) y+(3-k) z+(5 k-4)=0 \quad \text{...(i)}$
D.R.' $s$ of normal of plane (i) are
$$
= < (1+2 k),(2+k),(3-k)>
$$
Given, $5 x+3 y+6 z+8=0 \quad \text{...(ii)}$
D.R.'s of plane (ii) are $ < 5,3,6>$.
Since Eq. (i) is Perpendicular to the plane (ii),
$$
\begin{aligned}
&\therefore \quad 5(1+2 k)+(2+k) 3+6(3-k)=0 \\
&\Rightarrow \quad 5+10 k+6+3 k+18-6 k=0 \\
&\Rightarrow 7 k+29=0 \Rightarrow k=\frac{-29}{7}
\end{aligned}
$$
$\therefore$ Required equation of plane is
$$
\begin{aligned}
&(x-2 y+3 z-4)+\left(-\frac{29}{7}\right)(2 x+y-z+5)=0 \\
&\Rightarrow 7 x+14 y+21 z-28-58 x-29 y+29 z-145=0 \\
&\Rightarrow-51 x-15 y+50 z-173=0 \\
&\Rightarrow 51 x+15 y-50 z+173=0
\end{aligned}
$$
Hence, the correct answer is option: 'B'.