Let P be a polynomial in $\Bbb{RP}^2$. (It doesn't have to be in 2 dimensions, but assume it for clarity and simplicity)
For example: $P=3x^2+6y^2-7xy+10y-1$
Now, let's perform a projective transformation, $f$, on it.
$$f(x,y)=\left(\frac{a_1x+b_1y+c_1}{a_3x+b_3y+c_3},\frac{a_2x+b_2y+c_2}{a_3x+b_3y+c_3}\right)$$
Then it turns out that $f(P)=Q$ where Q is a polynomial of the same degree:
$$deg(P)=deg(Q)$$
I completely understand the cases where $deg(P)\le2$.
I have no idea why this is true when the degree is larger than 2. I'd appreciate a clear explanation for why this happens (if it does).
Writing your example in homogeneous coordinates, we get
$$P(x,y,z) = 3x^2 + 6y^2 - 7xy + 10yz - 1z^2$$
which has equal degree in all terms. The transformation in homogeneous coordinates would be
$$f:v\mapsto Mv=\begin{pmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{pmatrix}v$$
Now if you plug a transformed point into your original curve, you have
$$P(M\cdot(x,y,z)^T) = 3(a_1x + b_1y + c_1z)^2 + 6(a_2x+b_2y+c_2z)^2 - 7(a_1x + b_1y + c_1z)(a_2x+b_2y+c_2z) + \cdots$$
Now every term of the original curve becomes a constant times a product of parentheses, where the number of parentheses (taking powers into account) equals the degree. Expanding them using the distributive law you get a homogeneous polynomial whose degree again equals the original degree. So degree is preserved by all of this.
Well, plugging the transformed point into the original curve is essentially the wrong direction: it will give you a description of the inversely transformed curve. But to fix that, you simply invert the transformation matrix, without affecting the degree arguments stated above.