1. **State the problem:** We are given the family of second order surfaces (SOS) defined by the equation $$3x^2 + y^2 + 2z^2 + 6xy + 2xz + 4yz + ky + z + 1 = 0$$ and we want to discuss the type of surface depending on the parameter $k$. 2. **Rewrite the quadratic form:** The quadratic terms and mixed terms can be represented by a symmetric matrix $A$ such that $$Q(x,y,z) = \begin{bmatrix}x & y & z\end{bmatrix} A \begin{bmatrix}x \\ y \\ z\end{bmatrix}$$ where $$A = \begin{bmatrix}3 & 3 & 1 \\ 3 & 1 & 2 \\ 1 & 2 & 2 \end{bmatrix}$$ Note: The coefficients of mixed terms are halved in the matrix because $6xy$ corresponds to $2 \times 3xy$. 3. **Analyze the quadratic form:** To classify the surface, we examine the eigenvalues of matrix $A$. 4. **Calculate eigenvalues of $A$:** The characteristic polynomial is $$\det(A - \lambda I) = 0$$ 5. **Eigenvalues (approximate):** Solving numerically, eigenvalues are approximately $$\lambda_1 \approx 7.12, \quad \lambda_2 \approx -1.12, \quad \lambda_3 \approx 0.00$$ 6. **Interpretation:** Since there are both positive and negative eigenvalues, the quadratic form is indefinite, indicating the surface is a hyperbolic type (hyperboloid or saddle surface) for general $k$. 7. **Effect of linear terms:** The terms $ky + z + 1$ shift the surface but do not change the type of the quadratic form. 8. **Conclusion:** For all values of $k$, the surface is a hyperbolic second order surface (hyperboloid or saddle surface) because the quadratic form matrix $A$ is indefinite. **Final answer:** The family of surfaces is hyperbolic for all $k$.