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  4. Let the line (x-2/3) = (y-1/-5) = (z+2/2) lie in the plane x + 3y-α z + β = 0. Then (α, β) equals

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Q. Let the line $ \frac{x-2}{3} = \frac{y-1}{-5} = \frac{z+2}{2}$ lie in the plane $x + 3y-\alpha z + \beta = 0$. Then $\left(\alpha,\, \beta\right)$ equals

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Solution:

$ \because$ The line $\frac{x-2}{3} = \frac{y-1}{-5} = \frac{z+2}{2}$ lie in the plane
$x + 3y-\alpha z + \beta = 0$
$\therefore Pt \left(2, 1, - 2\right)$ lies on the plane i.e. $2 + 3 + 2\alpha + \beta = 0$
or $2\alpha + \beta + 5 =0\quad....\left(i\right)$
Also normal to plane will be perpendicular to line,
$\therefore 3\cdot 1-5\times3 + 2\times \left(-\alpha\right) = 0 \Rightarrow \alpha = -6$
From equation $\left(i\right)$ then, $\beta = 7$
$\therefore \left(\alpha,\,\beta\right) = \left(- 6, 7\right)$