If $\left[\begin{array}{lll}1 & 3 & 3 \\ 1 & 4 & 4 \\ 1 & 3 & 4\end{array}\right]\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\left[\begin{array}{l}12 \\ 15 \\ 13\end{array}\right]$, then the value of $x^2+y^2+z^2=$

The matrix equation corresponds to the system:

$x + 3y + 3z = 12$

$x + 4y + 4z = 15$

$x + 3y + 4z = 13$

Subtracting the first equation from the second gives: $y + z = 3$.

Subtracting the first from the third yields: $z = 1$.

Substituting $z = 1$ into $y + z = 3$ gives $y = 2$.

Then, from the first equation: $x + 3(2) + 3(1) = 12$, so $x = 3$.

Thus, $x^{2} + y^{2} + z^{2} = 3^{2} + 2^{2} + 1^{2} = 9 + 4 + 1 = 14$.

Final answer: $\boxed{D}$