If \(A X=B\), where \(A=\left[\begin{array}{lll}1 & 3 & 3 \\ 1 & 4 & 4 \\ 1 & 3 & 4\end{array}\right], X=\left[\begin{array}{l}x \\ y \\ z\end{array}\right]\) and \(B=\left[\begin{array}{l}12 \\ 15 \\ 13\end{array}\right]\), then \(x^{2}+y^{2}+z^{2}=\)

(A)
Given \(\mathrm{AX}=\mathrm{B}\)
\(\left[\begin{array}{lll}
1 & 3 & 3 \\
1 & 4 & 4 \\
1 & 3 & 4
\end{array}\right]\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]=\left[\begin{array}{c}
12 \\
15 \\
13
\end{array}\right]\)
\(\mathrm{R}_{2} \rightarrow \mathrm{R}_{2}-\mathrm{R}_{1}\) and \(\mathrm{R}_{3} \rightarrow \mathrm{R}_{3}-\mathrm{R}_{1}\)
\(\left[\begin{array}{lll}1 & 3 & 3 \\ 0 & 1 & 1 \\ 0 & 0 & 1\end{array}\right]\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\left[\begin{array}{l}12 \\ 3 \\ 1\end{array}\right] \Rightarrow\left[\begin{array}{l}x+3 y+3 z \\ y+z \\ z\end{array}\right]=\left[\begin{array}{l}12 \\ 3 \\ 1\end{array}\right]\)
\(\begin{aligned}
\therefore & x+3 y+3 z=12 \\
& y+z=3 \\
& z=1
\end{aligned}\)
Thus \(z=1, y=2, x=3\)
\(\therefore x^{2}+y^{2}+z^{2}=9+4+1=14\)