$latex \mathbf{12}$
$latex a : b : c = 2 : 3 : 4$
$latex \therefore \frac{a}{2} = \frac{b}{3} = \frac{c}{4} = k (let)&s=1$
$latex => a = 2k, b = 3k, c = 4k$
Given, $latex 2a – 3b + 4c = 33$
$latex => 2\times 2k – 3\times 3k + 4\times 4k = 33$
$latex => 4k – 9k + 16k = 33$
$latex => 11k = 33$
$latex => k = \frac{33}{11} = 3&s=1$
$latex \therefore c = 4k = 4\times 3 = 12$
Hence option [C] is correct answer.