= Solution
Use part (a) to put the four <projective points> at
$$
[1:0:0],\quad[0:1:0],\quad[0:0:1],\quad[1:1:1].
$$
A <quadratic polynomial> vanishing at the first three has no square terms, and vanishing at the fourth gives
$$
q=Axy+Bxz+Cyz,\qquad A+B+C=0.
$$
These form a two-dimensional <vector space>. Choose the two members
$$
q_0=xy+xz-2yz,\qquad q_1=xy-2xz+yz.
$$
Their coefficient triples are independent and both satisfy the relation. Their symmetric <bilinear form> matrices are of the form
$$
B(A,B,C)=\frac12\begin{pmatrix}0&A&B\\A&0&C\\B&C&0\end{pmatrix},\qquad
\det B(A,B,C)=\frac{ABC}{4}.
$$
For $(A,B,C)=(1,1,-2)$ and $(1,-2,1)$, both determinants are $-1/2$, so both selected conics are nonsingular. Any coefficient triple with zero sum is uniquely a <linear combination> of these two. Hence every conic through the four points has matrix
$$
\boxed{\lambda B_0+\mu B_1,\qquad(\lambda,\mu)\ne(0,0).}
$$
Undoing the <projective transformation> carries the entire family back to the original four points, preserving nonsingularity. This is a <pencil of conics through four points>.
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