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Let the line $\frac{x-2}{3}=\frac{y-1}{-5}=\frac{z+2}{2}$ lie in the plane $x+3 y-\alpha z+\beta=0$. Then $(\alpha, \beta)$ equals
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Verified Answer
The correct answer is:
$(-6,7)$
$\because$ The line $\frac{x-2}{3}=\frac{y-1}{-5}=\frac{z+2}{2}$ lie in the plane
$$
x+3 y-\alpha z+\beta=0
$$
$\therefore \operatorname{Pt}(2,1,-2)$ lies on the plane i.e. $2+3+$ $2 \alpha+\beta=0$
or $2 \alpha+\beta+5=0$
$\ldots .(\mathrm{i})$
Also normal to plane will be perpendicular to line,
$\therefore 3 \times 1-5 \times 3+2 \times(-\alpha)=0 \Rightarrow \alpha=-6$
From equation (i) then, $\beta=7$
$\therefore(\alpha, \beta)=(-6,7)$
$$
x+3 y-\alpha z+\beta=0
$$
$\therefore \operatorname{Pt}(2,1,-2)$ lies on the plane i.e. $2+3+$ $2 \alpha+\beta=0$
or $2 \alpha+\beta+5=0$
$\ldots .(\mathrm{i})$
Also normal to plane will be perpendicular to line,
$\therefore 3 \times 1-5 \times 3+2 \times(-\alpha)=0 \Rightarrow \alpha=-6$
From equation (i) then, $\beta=7$
$\therefore(\alpha, \beta)=(-6,7)$
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