MathematicsMedium249×since 2002Q2391Let the line \,\,\,\,\, x−23=y−1−5=z+22{{x - 2} \over 3} = {{y - 1} \over { - 5}} = {{z + 2} \over 2}3x−2=−5y−1=2z+2 lie in the plane \,\,\,\,\, x+3y−αz+β=0.x + 3y - \alpha z + \beta = 0.x+3y−αz+β=0. Then (α,β)\left( {\alpha ,\beta } \right)(α,β) equalsA(−6,7)(-6,7)(−6,7)B(5,−15)(5,-15)(5,−15)C(−5,5)(-5,5)(−5,5)D(6,−17)(6, -17)(6,−17)Check answerSkip