Kernel of the quadratic form $q(x,y,z)=3x^2+6xy+10xz+2yz+3z^2$ Let $q:\mathbb{R^3} \to \mathbb{R}$ such that $$q(x,y,z)=3x^2+6xy+10xz+2yz+3z^2.$$ I have to determine rank, signature, kernel, and the canonic form of q (with its matrix). I have solved most of the problem; it only remains to find the kernel, but weidly it turns out to be two-dimensional; probably, I got wrong somehow. Could you show me the correct reasoning to find the kernel of a quadratic form?

You should have got without problems that $$q(x,y,z) = \begin{pmatrix} x & y & z \end{pmatrix}\begin{pmatrix} 3 & 3 & 5 \\\ 3 & 0 & 1 \\\ 5 & 1 & 3\end{pmatrix}\begin{pmatrix}x \\\ y \\\ z\end{pmatrix}.$$ Call $Q$ that middle matrix. Note that $\det Q = -3 -3(4)+5(3) = 0$, but the minor $$\begin{pmatrix} 3 & 3 \\\ 3 & 0 \end{pmatrix}$$has non-zero determinant, so $Q$ has rank $2$. I won't find a basis for the kernel here, but by rank-nullity theorem: $$3 = \dim \ker Q + {\rm rank \ }Q \implies \dim \ker Q = 1,$$ so there must be something wrong with your calculations. It is easier to check it for you if you add your work as an edit in the question. I won't comment on signature and canonical form since you managed that part without issues.