Question
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 :
A.
$$\left( { - 6,\,7} \right)$$
B.
$$\left( {5,\, - 15} \right)$$
C.
$$\left( { - 5,\,5} \right)$$
D.
$$\left( {6,\, - 17} \right)$$
Answer :
$$\left( { - 6,\,7} \right)$$
Solution :
$$\because $$ The line $$\frac{{x - 2}}{3} = \frac{{y - 1}}{{ - 5}} = \frac{{z + 2}}{2}$$ lies 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 $$2\alpha + \beta + 5 = 0.....({\text{i}})$$
Also normal to plane will be perpendicular to line,
$$\eqalign{
& \therefore 3 \times 1 - 5 \times 3 + 2 \times \left( { - \alpha } \right) = 0 \cr
& \Rightarrow \alpha = - 6 \cr} $$
From equation (i) then, $$\beta = 7$$
$$\therefore \,\,\left( {\alpha ,\,\beta } \right) = \left( { - 6,\,7} \right)$$