If the line x−$$23=y$$−$$1$$−$$5=z+22$$ lie in the plane $$x+3y$$−α$$z+$$β$$=0$$ then $$12$$α$$2+$$β$$2=$$

(2$$,$$1$$,−$$2) lies $$x+3y$$−α$$z+$$β$$=0$$⇒$$2$$α$$+$$β$$=$$−$$5$$ Line lies in the plane ⇒$$3(1)+($$−$$5)(3)+2($$−α$$)=0$$⇒$$2$$α$$=$$−$$12$$⇒α$$=$$−$$6$$,β$$=7$$α$$2+$$β$$22=36+492=42.5$$

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Question

$If the line \frac{x - 2}{3} = \frac{y - 1}{- 5} = \frac{z + 2}{2} lie in the plane x + 3 y - α z + β = 0 then \frac{1}{2} (α^{2} + β^{2}) =$

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Solution

$\begin{array}{l} (2, 1, - 2) lies x + 3 y - α z + β = 0 \\ \Rightarrow 2 α + β = - 5 \\ Line lies in the plane \\ \Rightarrow 3 (1) + (- 5) (3) + 2 (- α) = 0 \Rightarrow 2 α = - 12 \\ \Rightarrow α = - 6, β = 7 \\ \frac{α^{2} + β^{2}}{2} = \frac{36 + 49}{2} = 42.5 \end{array}$

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