Explanation

(c)
Let $f(x) = a_{0}x^{2} + a_{1}x + a_{2}$
and $g(x) = b_{2}x^{2} + b_{1}x + b_{2}$
Also, $h(x) = c_{0}x^{2} + c_{1}x + c_{2}$
Then, $\Delta(x) = \left| \begin{matrix}
f(x) & g(x) & h(x) \\
2a_{0}x + a_{1} & 2b_{0}x + b_{1} & 2c_{0}x + c_{1} \\
2a_{0} & 2b_{0} & 2c_{0}
\end{matrix} \right|$
$= x\left| \begin{matrix}
f(x) & g(x) & h(x) \\
2a_{0} & 2b_{0} & 2c_{0} \\
2a_{0} & 2b_{0} & 2c_{0}
\end{matrix} \right| + \left| \begin{matrix}
f(x) & g(x) & h(x) \\
a_{1} & b_{1} & c_{1} \\
2a_{0} & 2b_{0} & 2c_{0}
\end{matrix} \right|$
$= 0 + 2\left| \begin{matrix}
f(x) & g(x) & h(x) \\
a_{1} & b_{1} & c_{1} \\
a_{0} & b_{0} & c_{0}
\end{matrix} \right|$
$= 2\lbrack\left( b_{1}c_{0} - b_{0}c_{1} \right)f(x) - \left( a_{1}c_{0} - a_{0}c_{1} \right)g(x) + \left( a_{1}b_{0} - a_{0}b_{1} \right)h(x)\rbrack$
Hence, degree of $\Delta(x) \leq 2$