Answer
$$B^TC^T-(CB)^T= \left[\begin{array}{r}{0}& {0}&{0}\\{0}&{0}&{0}\\{0}&{0}&{0}\end{array}\right].$$
Work Step by Step
Since $$
B=\left[\begin{array}{ll}{4} & {-2}&{1}\\
{0}&{2}&{3}\end{array}\right], \quad C=\left[\begin{array}{r}{1}&{2} \\ {3}&{4}\\{5}&{6}\end{array}\right]
$$
we have
$$
B^T=\left[\begin{array}{r}{4}&{0} \\ {-2}&{2}\\{1}&{3}\end{array}\right], \quad C^T=\left[\begin{array}{r}{1}& {3}&{5}\\{2}&{4}&{6}\end{array}\right]
$$ and
$$B^TC^T=\left[\begin{array}{r}{4}&{0} \\ {-2}&{2}\\{1}&{3}\end{array}\right] \left[\begin{array}{r}{1}& {3}&{5}\\{2}&{4}&{6}\end{array}\right]=\left[\begin{array}{r}{4}& {12}&{20}\\{2}&{2}&{2}\\{7}&{15}&{23}\end{array}\right].$$
Moreover, we get that
$$CB=\left[\begin{array}{r}{1}&{2} \\ {3}&{4}\\{5}&{6}\end{array}\right]\left[\begin{array}{ll}{4} & {-2}&{1}\\
{0}&{2}&{3}\end{array}\right]=\left[\begin{array}{r}{4}& {2}&{7}\\{12}&{2}&{15}\\{20}&{2}&{23}\end{array}\right]$$
and hence $$(CB)^T=\left[\begin{array}{r}{4}& {12}&{20}\\{2}&{2}&{2}\\{7}&{15}&{23}\end{array}\right].$$
It is easy to note that
$$B^TC^T-(CB)^T=\left[\begin{array}{r}{4}& {12}&{20}\\{2}&{2}&{2}\\{7}&{15}&{23}\end{array}\right]-\left[\begin{array}{r}{4}& {12}&{20}\\{2}&{2}&{2}\\{7}&{15}&{23}\end{array}\right]=\left[\begin{array}{r}{0}& {0}&{0}\\{0}&{0}&{0}\\{0}&{0}&{0}\end{array}\right].$$
