• # question_answer The points $A(-1,3,0)$, $B\,(2,\,2,\,1)$ and $C\,(1,\,1,\,3)$ determine a plane. The distance from the plane to the point $D(5,\,7,8)$ is                                                               [AMU 2001] A)            $\sqrt{66}$ B)            $\sqrt{71}$ C)            $\sqrt{73}$ D)            $\sqrt{76}$

Equation of plane passing through $(-1,\,3,\,0)$ is $A(x+1)+B(y-3)+C(z-0)=0$                     ......(i)            Also, plane (i) is passing through the points $(2,\,2,\,1)$ and (1, 1, 3).  So,       $3A-B+C=0$                     .....(ii)                                 $2A-2B+3C=0$                .....(iii)            Solving (ii) and (iii), $\frac{A}{-3+2}=\frac{B}{2-9}=\frac{C}{-6+2}$            \ $A:B:C=-1:-7:-4$ or $A:B:C=1:7:4$            From (i), $1(x+1)+7(y-3)+4(z)=0$            or $x+7y+4z-20=0$            $\therefore$Distance from the plane to the point (5, 7, 8) is,                  $\frac{1\times 5+7\times 7+4\times 8-20}{\sqrt{{{1}^{2}}+{{7}^{2}}+{{4}^{2}}}}=\frac{5+49+32-20}{\sqrt{66}}=\frac{66}{\sqrt{66}}=\sqrt{66}$.