If the equation of the plane passing through the line of intersection of the planes 2 x-y+…

Question

If the equation of the plane passing through the line of intersection of the planes $2 x-y+z=3,4 x-3 y+5 z+9=0$ and parallel to the line $$\frac{x+1}{-2}=\frac{y+3}{4}=\frac{z-2}{5}$$ is $a x+b y+c z+6=0$, then $a+b+c$ is equal to :

Accepted Answer

Equation of plane intersection of two plane

$$
\mathrm{P}_1+\lambda \mathrm{P}_2=0
$$

$$
\mathrm{P}_1: 2 x-y+z=3, \text { and } \mathrm{P}_2: 4 x-3 y+5 z+9=0
$$

Equation of any plane passing through the intersection of given planes is

$(2 x-y+z-3)+\lambda(4 x-3 y+5 z+9)=0$

$$
\Rightarrow(2+4 \lambda) x+(-1-3 \lambda) y+(1+5 \lambda) z+(-3+9 \lambda)=0
$$ ..........(i)

Plane (i) is parallel to given line

$$
\begin{aligned}
& \therefore  (-2)(2+4 \lambda)+4(-1-3 \lambda)+5(1+5 \lambda)  =0 \\
& \Rightarrow  -4-8 \lambda-4-12 \lambda+5+25 \lambda =0 \\
& \Rightarrow  5 \lambda-3 =0 \\
& \Rightarrow  \lambda =\frac{3}{5}
\end{aligned}
$$

Putting the value of $\lambda$ in Eq. (i), we get

$$
\begin{aligned}
& (2 x-y+z-3)+\frac{3}{5}(4 x-3 y+5 z+9)=0 \\
& \Rightarrow 11 x-7 y+10 z+6=0 \\
& \therefore a=11, b=-7, c=10 \text { and } d=6 \\
& \therefore a+b+c=11-7+10=14
\end{aligned}
$$

Solutions · 1