= Solution
A <blowup of an algebraic variety> supplies a reduced special fibre with all three requested defects. Let $B=\mathbb A^3$ have coordinates $(x,y,z)$, let $p=(0,1,0)$, and let $\pi:X=\operatorname{Bl}_pB\to B$. Take $Y=\mathbb A^1$ and the <regular map>
$$
f:X\longrightarrow Y,\qquad f=(xy)\circ\pi.
$$
For an explicit construction, $X$ is the closed <subvariety> of $\mathbb A^3\times\mathbb P^2$, with projective coordinates $[u:v:w]$, defined by
$$
xv=(y-1)u,\qquad xw=zu,\qquad (y-1)w=zv.
$$
These equations say that $(x,y-1,z)$ is proportional to $(u,v,w)$. The three standard projective charts are <affine spaces> $\mathbb A^3$, so $X$ is a <smooth variety> which is an <irreducible variety> embedded in a <quasi-projective algebraic set>. Its <algebraic exceptional divisor> is $E=\pi^{-1}(p)\cong\mathbb P^2$.
If $t\ne0$, the surface $xy=t$ misses $p$, and the <blowup of an algebraic variety> is an <isomorphism> over it. Hence
$$
\boxed{X_t\cong\operatorname{Spec}k[x,x^{-1},z]\cong\mathbb G_m\times\mathbb A^1.}
$$
It is a <smooth variety>, an <affine variety>, an <irreducible variety>, and has <algebraic dimension> $2$.
For $t=0$, the <fibre of a morphism> has three <irreducible components>: the <strict transform of an algebraic subvariety> $\widetilde{V(x)}$, the unchanged plane $V(y)$, and the <algebraic exceptional divisor> $E\cong\mathbb P^2$. The centre $p$ lies on $V(x)$ but not on $V(y)$; this is why all three components have multiplicity one. To verify reducedness and the crossing directly near $E$, use the chart $v\ne0$, putting
$$
a=x/(y-1),\qquad b=y-1,\qquad c=z/(y-1).
$$
Here $x=ab$, $y=1+b$, $z=bc$, and $f=ab(1+b)$. Near $b=0$, the factor $1+b$ is a <unit>, so the <fibre of a morphism> is $ab=0$, a reduced union of two crossing planes. The $u$-chart gives $f=x(1+xv)$, whose second factor is a <unit> along $E$; the $w$-chart gives $f=zu(1+zv)$ and the same reduced crossing. Away from $E$, reducedness follows from $xy=0$ in $B$. Thus the scheme-theoretic <fibre of a morphism> is itself reduced.
It has <singular points of an algebraic variety> already along $V(x,y)$, which misses $p$: locally its equation is $xy=0$, with <Zariski tangent space> of <dimension of a vector space> $3$ but <local dimension of an algebraic variety> $2$. Finally, $E$ is a closed <subvariety> of $X_0$. A closed <subvariety> of an <affine variety> is an <affine variety>, whereas $\mathbb P^2$ is not: all its <global regular functions> are constant, which cannot be the <coordinate ring> of a positive-dimensional <affine variety>. Consequently \b[$X_0$ is singular, reducible, and nonaffine], as required. Choosing the centre off the intersection of the original planes avoids introducing a nonreduced special <fibre of a morphism>.
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