If the line $\frac{x-2}{3}=\frac{y-1}{-5}=\frac{z+2}{2}$ lies in the plane $x+3 y-\alpha z+\beta=0$, then $(\alpha, \beta)=$

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Aicharya · Accepted Answer

Given equation of line $\frac{x-2}{3}=\frac{y-1}{-5}=\frac{z+2}{2}$ $\therefore \quad$ The line passes through $(2,1,-2)$ The above point lies on the plane $\begin{aligned} & x+3 y-\alpha z+\beta=0 \\ & \Rightarrow 2+3+2 \alpha+\beta=0 \\ & \Rightarrow 2 \alpha+\beta=-5 \end{aligned}$
Also the given line is perpendicular to the normal to the plane $\begin{aligned} & a_1 a_2+b_1 b_2+c_1 c_2=0 \\ & \Rightarrow 3(1)+(-5)(3)-2(\alpha)=0 \\ & \Rightarrow \alpha=-6 \end{aligned}$ $\begin{aligned} & \text { From (i) } \\ & \beta=7 \\ \therefore \quad & \alpha \beta=-42 \\ & (\alpha, \beta)=(-6,7)\end{aligned}$

If the line $\frac{x-2}{3}=\frac{y-1}{-5}=\frac{z+2}{2}$ lies in the plane $x+3 y-\alpha z+\beta=0$, then…

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