Đáp án:

Câu `19.`

`vec(MC)=(-2-a;3-b;3-c)`

`vec(AB)=(1;-3;4)`

Vì `ABCM` là hình bình hành `<=>vec(MC)=vec(AB)`

`<=>{(-2-a=1),(3-b=-3),(3-c=4):}`

`<=>{(a=-3),(b=6),(c=-1):}`

`=>P=a^2+b^2-c^2=9+36-1=44`

Câu `20.`

Theo đề:

`vec(AD)=(2;0;0)=2veci`

`vec(AB)=(0;2;0)=2vecj`

`vec(A A')=(0;0;2)=2veck`

`@`Ta có: `vec(AC')=vec(AD)+vec(AB)+vec(A A')` (qt hình hộp)

`=2veci+2vecj+2veck`

`=>{(a=2),(b=2),(c=2):}`

`=>a+b+c=6`

Câu 19:

`\vec{AB} = (1;-3;4)`

`\vec{MC} = (- 2 - a;3 - b; 3 - c)`

Ta có: `\vec{AB} = \vec{MC}`

`<=> {(- 2 - a  = 1),(3 - b = - 3),(3 - c = 4):}`

`<=> {(a = - 3),(b = 6),(c = - 1):}`

`=> P = a^{2} + b^{2} - c^{2} = 44`

Câu 20: `A \equiv O (0;0;0)`

`AD \in Ox => AD = (2;0;0) => AD = 2\vec{i}`

`AB \in Oy => AB = (0;2;0) => AD = 2\vec{j}`

`AA' \in Oz => AD = (0;0;2) => AD = 2\vec{k}`

Ta có: `\vec{AC'} = \vec{AC} + \vec{CC'}`

`= \vec{AD} + \vec{AB} + \vec{AA'}`

`= 2\vec{i} + 2\vec{j} + 2\vec{k}`

`=> a = b = c = 2`

Vậy `a + b + c = 6`

$\color{#FA8072}{\text{$\textit{$\circ$ hungnguyen4269}$}}$