Find the base of the right triangle

Since $\sin \theta = \frac{12}{13}$, and in a right triangle $\text{Perpendicular}^2 + \text{Base}^2 = \text{Hypotenuse}^2$, we have $\text{Base} = \sqrt{13^2 - 12^2} = \sqrt{169 - 144} = \sqrt{25} = 5$.

$\text{Base} = 5$

Calculate sec θ

The value of $\cos \theta = \frac{\text{Base}}{\text{Hypotenuse}} = \frac{5}{13}$. Since $\sec \theta = \frac{1}{\cos \theta}$, it follows that $\sec \theta = \frac{13}{5}$.

$\sec \theta = \frac{13}{5}$

Compute the final expression

Substitute the values into $26 \sin \theta - 10 \sec \theta$: $26 \times (\frac{12}{13}) - 10 \times (\frac{13}{5})$.

$26 \times \frac{12}{13} - 10 \times \frac{13}{5} = 2 \times 12 - 2 \times 13 = 24 - 26 = -2$