= Solution
Write $a=t_1$, $b=t_2$, $c=t_3$, and $A=\mathbb C[a,b,c]/(ac-b^2)$. It is the <quadratic cone invariant ring>
$$
A\cong\mathbb C[u^2,uv,v^2]=\mathbb C[u,v]^{\{\pm1\}},
$$
where the involution negates both $u$ and $v$. The invariant monomials have even total degree and are generated by $u^2,uv,v^2$; reduction by $ac=b^2$ proves the asserted presentation.
To verify the <normal variety> property, suppose $q\in\operatorname{Frac}(A)$ is integral over $A$. The same monic equation makes $q$ integral over $\mathbb C[u,v]$. This <polynomial ring> is an <integrally closed domain>, so $q\in\mathbb C[u,v]$. Since $q$ belongs to the <fraction field> of $A$, it is invariant under the involution, and hence belongs to $A$. Thus \b[$X$ is normal]. The <gradient> of its equation is $(c,-2b,a)$, so the <Jacobian criterion> shows that the origin $o$ is its only <singular point of an algebraic variety>. In particular \b[$X$ is not smooth].
Take the <prime Weil divisor> $D=V_X(a,b)$ and $x=o$. We show that $D$ is not a <Cartier divisor> at $o$. In the <local ring>
$$
R=A_{(a,b,c)},\qquad\mathfrak m=(a,b,c)R,
$$
its prime <ideal> is $I=(a,b)R$. The quotient $I/\mathfrak mI$ has <dimension of a vector space> $2$ over $\mathbb C$: $a,b$ generate it, and they are independent since $\mathfrak mI\subset\mathfrak m^2$ while the relation $ac-b^2$ has no linear term. By <Nakayama lemma>, $I$ cannot be generated by one element.
On a <normal variety>, the ideal of an effective <prime Weil divisor> is the <divisorial ideal> of functions with order at least one along that divisor and order at least zero along every other <prime Weil divisor>. If $D$ were a <Cartier divisor> at $o$, a local defining <rational function> $h$ would identify this ideal with $hR$, because an <integrally closed domain> is the intersection of its height-one <localizations at a prime ideal> inside its <fraction field>. That would make $I$ principal, contradicting the calculation. Therefore $D$ has a nonzero class in the local <divisor class group>.
Now let $D'=D+\operatorname{div}(q)$ be any <linearly equivalent Weil divisor>. If $o$ were outside its <support of a Weil divisor>, there would be an <open subset> containing $o$ on which $D'$ is zero. On that neighbourhood $D=-\operatorname{div}(q)$ would be a <principal Weil divisor>, hence a <Cartier divisor>, which is impossible. Therefore \b[this point is unavoidable in every representative]:
$$
\boxed{o\in\operatorname{Supp}(D')\quad\text{for every }D'\sim D.}
$$
The argument does not assume that $D'$ is effective. For comparison, $\operatorname{div}(a)=2D$ because on the chart $c\ne0$, $a=b^2/c$; the obstruction has order two, as in the <Divisor class group of an A-type surface singularity>.
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