Let \(\cfrac{a}{2} = \cfrac{b}{3} = \cfrac{c}{4} = \cfrac{2a - 3b + 4c}{p} = k\) Therefore, \(a = 2k\), \(b = 3k\), and \(c = 4k\) Again, \(\cfrac{2a - 3b + 4c}{p} = k\) i.e., \(\cfrac{2 \times 2k - 3 \times 3k + 4 \times 4k}{p} = k\) i.e., \(\cfrac{4k - 9k + 16k}{p} = k\) i.e., \(\cfrac{11k}{p} = k\) i.e., \(pk = 11k\) i.e., \(p = 11\) (Answer)