Answer:
x = 2 , y = 1 , z = 2
Step-by-step explanation:
[tex]\overrightarrow{a} \perp \overrightarrow{c} \Longrightarrow \left( \begin{gathered}1\\ 0\\ -1\end{gathered} \right) \cdot \left( \begin{gathered}x\\ y\\ z\end{gathered} \right) =0 \Longleftrightarrow x - z=0 \Longleftrightarrow x = z[/tex]
[tex]\overrightarrow{b} \perp \overrightarrow{c} \Longrightarrow \left( \begin{gathered}-2\\ 2\\ 1\end{gathered} \right) \cdot \left( \begin{gathered}x\\ y\\ z\end{gathered} \right) =0 \Longleftrightarrow -2x +2y +z=0[/tex]
Now ,we have to solve the system:
[tex]\begin{cases}x=z&\\ -2x+2y+z=0&\end{cases}[/tex]
[tex]\Longleftrightarrow \begin{cases}x=z&\\ -x+2y=0&\end{cases}[/tex]
[tex]\Longleftrightarrow \begin{cases}x=z&\\ x=2y&\end{cases}[/tex]
then x = 2y = z
[tex]\Longrightarrow \overrightarrow{c} \left( \begin{gathered}2y\\ y\\ 2y\end{gathered} \right)[/tex]
|C| = 3 ⇒ (2y)² + y² + (2y)² = 9 ⇒ 9y² = 9 ⇒ y² = 1 ⇒ y = ±1 ⇒ y = 1
(x>0 ⇒ y>0)
[tex]\Longrightarrow \overrightarrow{c} \left( \begin{gathered}2\\ 1\\ 2\end{gathered} \right)[/tex]