Show that the (Veronese-like) surface is given by zero locus of the following polynomials

Question

Consider the following set in $\mathbb R^6$: $$ S= \bigl \{(x_1^2,x_2^2,x_3^2,x_1x_2,x_2x_3,x_3x_1) \mid (x_1,x_2,x_3) \in \mathbb R^3, \; x_1^2+x_2^2+x_3^2 = 1 \bigr\}. $$ If we denote by $(y_1,y_2,y_3,y_4,y_5,y_6)$ the coordinates on $\mathbb R^6$ then we can see that on $S$ the following relations hold: $$ y_1 y_2 = y_4^2, \quad y_2 y_3 = y_5^2, \quad y_3 y_1 = y_6^2, \\ y_1 y_5 = y_4 y_6, \quad y_2 y_6 = y_4 y_5, \quad y_3 y_4 = y_5 y_6, \\ y_1+y_2+y_3 = 1. $$ Thus $S$ lies in the common zero-locus of 6 degree-two polynomals and one degree-one polynomial on $\mathbb R^6$. It can be shown also that all the polynomials are linearly independent over $\mathbb R$. But how to see the that their common zero-locus is exactly $S$?