= Solution
For an <algebraically closed field> $k$, the strong <Hilbert Nullstellensatz> says that every <ideal> $J\subseteq k[x_1,\ldots,x_n]$ satisfies
$$
\boxed{I(V(J))=\sqrt J.}
$$
Its weak form says that $V(J)=\varnothing$ exactly when $J=(1)$. Now let $S=\mathbb C[x_0,\ldots,x_n]$ and let $I\subseteq S$ be a <homogeneous ideal>. The affine zero set of $I$ is the <affine cone> over $V_+(I)$ together with the origin. Hence, when the projective zero set is empty, either
$$
\boxed{I=(1)}
$$
or the affine zero set is exactly the origin. In the latter case the <Hilbert Nullstellensatz> gives
$$
\boxed{\sqrt I=(x_0,\ldots,x_n),}
$$
the <irrelevant ideal of projective space>. Equivalently, some power $(x_0,\ldots,x_n)^N$ is contained in $I$. These are precisely the alternatives in the <Projective Nullstellensatz>.
Let $V\subseteq\mathbb P^3$ be a <smooth quadric surface>. After a <projective linear transformation>, the <Segre embedding> identifies
$$
V\cong\mathbb P^1\times\mathbb P^1.
$$
Choose distinct points $p,q\in\mathbb P^1$. The two fibres
$$
C_p=\mathbb P^1\times\{p\},
\qquad
C_q=\mathbb P^1\times\{q\}
$$
are the promised <disjoint curves on a smooth quadric surface>: each is a smooth <projective line>, and $C_p\cap C_q=\varnothing$. By the <Bézout theorem>, any two nonempty curves in the <projective plane> intersect. An <isomorphism of algebraic varieties> preserves intersections and takes <projective curve>[projective curves] to projective curves, so these disjoint curves prove
$$
\boxed{V\not\cong\mathbb P^2.}
$$
Next let $W$ be a <smooth projective curve> and let $f:W\dashrightarrow Y\subseteq\mathbb P^N$ be a <rational map of projective varieties>[rational map] to a <projective variety>. At a point $P$ outside its initial domain, write
$$
f=[f_0:\cdots:f_N]
$$
using elements of the <function field> $\mathbb C(W)$. The <local ring of a smooth algebraic curve> at $P$ is a <discrete valuation ring>. If $m=\min_i v_P(f_i)$, multiplying every coordinate by a function of valuation $-m$ makes all coordinates regular at $P$ and at least one a unit there. They therefore define a <morphism of algebraic varieties>[morphism] near $P$. Repeating this at the finitely many missing points proves the <extension of a rational map from a smooth projective curve>.
Smoothness is essential. Let $\nu:\mathbb P^1\to C$ be the <normalization of a nodal curve> for a rational nodal cubic $C$. The inverse is a rational map $C\dashrightarrow\mathbb P^1$ on the smooth locus. If it extended to a morphism $g:C\to\mathbb P^1$, then $g\circ\nu$ would agree with the identity on a <Zariski-open set>[dense Zariski-open subset] of $\mathbb P^1$ and hence everywhere. But the two distinct points $a,b\in\mathbb P^1$ above the node would satisfy
$$
a=(g\circ\nu)(a)=g(\nu(a))=g(\nu(b))=(g\circ\nu)(b)=b,
$$
a contradiction. This is the <failure of rational-map extension on a singular curve>.
Finally take the <Fermat cubic curve>
$$
F=x^3+y^3+z^3
$$
and define the <projective hypersurface>
$$
Z=\left\{([x:y:z],[s:t])\in\mathbb P^2\times\mathbb P^1:
sF+txyz=0\right\}.
$$
Because $F$ and $xyz$ have no common <irreducible polynomial>[irreducible factor], the bihomogeneous polynomial $sF+txyz$ is irreducible, so $Z$ is an <algebraic variety>. Let
$$
\pi:Z\longrightarrow\mathbb P^1,
\qquad
([x:y:z],[s:t])\longmapsto[s:t]
$$
be the second <coordinate projection>. Every homogeneous cubic in three variables has a projective zero over $\mathbb C$, so every fibre is nonempty and $\pi$ is a <surjective function>[surjective] <morphism of algebraic varieties>[morphism]. Over $p=[1:0]$ the fibre is the <Fermat cubic curve>, which is smooth and has <genus of a smooth plane curve>[genus]
$$
\frac{(3-1)(3-2)}2=1.
$$
Over $q=[0:1]$ the fibre is
$$
V(xyz)=V(x)\cup V(y)\cup V(z),
$$
the union of exactly three distinct <projective line>[projective lines], hence exactly three <irreducible component>[irreducible components]. This is the <cubic pencil with a triangular member> required by the question.
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