(3cos260° + 2cot230° – 5sin245°) = ?

As we know that,

$\sin 45^{\circ}=\frac{1}{\sqrt{2}}$

$\cos 60^{\circ}=\frac{1}{2}$

$\cot 30^{\circ}=\sqrt{3}$

By substituting these values, we get

$\left(3 \cos ^{2} 60^{\circ}+2 \cot ^{2} 30^{\circ}-5 \sin ^{2} 45^{\circ}\right)=3\left(\frac{1}{2}\right)^{2}+2(\sqrt{3})^{2}-5\left(\frac{1}{\sqrt{2}}\right)^{2}$

$=3\left(\frac{1}{4}\right)+2(3)-5\left(\frac{1}{2}\right)$

$=\frac{3}{4}+6-\frac{5}{2}$

$=\frac{3+24-10}{4}$

$=\frac{17}{4}$

Hence, the correct option is (c).