Q97 / 150 · MHT CET 2024 (10 May Shift 1) · Mathematics
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 $(\alpha, \beta)=$
- A$(6,-7)$
- Correct answerB$(-6,7)$
- C$(5,-15)$
- D$(-5,15)$
Solution
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 $\begin{aligned} & a_1 a_2+b_1 b_2+c_1 c_2=0 \\ & \Rightarrow 3(1)+(-5)(3)-2(\alpha)=0 \\ & \Rightarrow \alpha=-6 \end{aligned}$ $\begin{aligned} & \text { From (i) } \\ & \beta=7 \\ \therefore \quad & \alpha \beta=-42 \\ & (\alpha, \beta)=(-6,7)\end{aligned}$
Also the given line is perpendicular to the normal to the plane $\begin{aligned} & a_1 a_2+b_1 b_2+c_1 c_2=0 \\ & \Rightarrow 3(1)+(-5)(3)-2(\alpha)=0 \\ & \Rightarrow \alpha=-6 \end{aligned}$ $\begin{aligned} & \text { From (i) } \\ & \beta=7 \\ \therefore \quad & \alpha \beta=-42 \\ & (\alpha, \beta)=(-6,7)\end{aligned}$
Hence, the correct answer is option: 'B'.