$\vec{BC}=(-3;\ -2;\ -3);\ \vec{BD}=(-3;\ -8;\ -1) \Rightarrow \vec{n}_{(BCD)}=[\vec{BC},\ \vec{BD}]=(11;\ -3;\ -9)$
Gọi $M(x;\ y;\ z)$ vì $MC \perp (BCD)\Rightarrow \vec{MC} =k . \vec{n}_{(BDC)}$
$\Rightarrow \dfrac{x}{11}=\dfrac{y}{-3}=\dfrac{z}{-9}\Rightarrow y=\dfrac{-3}{11}x;\ z=-\dfrac{9}{11}x \ (1)$
Lại có $MC^2 = 100 = x^2 +(4-y)^2 +(z+1)^2 \ (2)$
Từ $(1);\ (2)$ là tìm được $M$