If equation of the plane that contains the point (-2,3,5) and is perpendicular to each of …

Question

If equation of the plane that contains the point $(-2,3,5)$ and is perpendicular to each of the planes $2 x+4 y+5 z=8$ and $3 x-2 y+3 z=5$ is $$\alpha x+\beta y+\gamma z+97=0$$ then $$\alpha+\beta+\gamma=$$

Accepted Answer

The equation of plane that passes through the point $(-2,3,5)$ is

$a(x+2)+b(y-3)+c(z-5)=0$ ..........(i)

The plane is perpendicular to

$$
\[\begin{gathered} 2 x+4 y+5 z  =8 \text { and } 3 x-2 y+3 z=5 \ \end{gathered}$$
\]

$$
\[\begin{gathered} \therefore   2 a+4 b+5 c=0 ..........(ii) \ \text { and }  3 a-2 b+3 c=0 ..........(iii) \end{gathered}$$
\]

$$
\begin{aligned}
& \therefore \quad \frac{\mathrm{a}}{\left|\[\begin{array}{cc} 4 & 5 \ -2 & 3 \end{array}\end{aligned}$$\right|}=\frac{-\mathrm{b}}{\left|$$\begin{array}{ll} 2 & 5 \ 3 & 3 \end{array}$$\right|}=\frac{\mathrm{c}}{\left|$$\begin{array}{cc} 2 & 4 \ 3 & -2 \end{array}$$\right|} \\
\& \Rightarrow \frac{\mathrm{a}}{22}=\frac{\mathrm{b}}{9}=\frac{\mathrm{c}}{-16}

\]

$\therefore$ From Equation (i)
equation of plane

$$
\begin{aligned}
&22(x+2)+9(y-3)-16(z-5)=0 \\
&\Rightarrow 22 x+9 y-16 z+97=0
\end{aligned}
$$

Here, $\alpha=22, \beta=9, \gamma=-16$

$$
\therefore \alpha+\beta+\gamma=22+9-16=15
$$

Solutions · 1