Cho $A(1;1;1), B(0;0;4), C(0;2;0)$. Viết phương trình mặt phẳng $(P)$ song song với mặt phẳng $(ABC)$ và cách $D(1;0;3)$ một khoảng bằng $\sqrt {6} $.
- Bước 1: Viết phương trình mặt phẳng (ABC) khi đã biết A,B,C: $x + 2y + z - 4 = 0$.
- Bước 2: mặt phẳng (P) song song với (ABC) nên phương trình mặt phẳng (P) có dạng: $x + 2y + z + d = 0   $
- Bước 3: Khoảng cách từ D đến (P) là: $\frac{{|4 + D|}}{{\sqrt {{1^2} + {2^2} + {1^2}} }} = \sqrt 6  \Leftrightarrow \left[ \begin{array}{l}
d = 2 \Rightarrow (P):x + 2y + z + 2 = 0\\
d = - 10 \Rightarrow (P):x + 2y + z - 10 = 0
\end{array} \right.$
Vậy có 2 mặt phẳng cần tìm.
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