(B) Let $p(x) = x^{3} - 2ax^{2} + ax - 1$.Since $(x-2)$ is a factor of $p(x)$,by the Factor Theorem,we must have $p(2) = 0$.Substituting $x = 2$ into

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(B) Let $p(x) = x^{3} - 2ax^{2} + ax - 1$.Since $(x-2)$ is a factor of $p(x)$,by the Factor Theorem,we must have $p(2) = 0$.Substituting $x = 2$ into the polynomial:$p(2) = (2)^{3} - 2a(2)^{2} + a(2) - 1 = 0$.$8 - 2a(4) + 2a - 1 = 0$.$8 - 8a + 2a - 1 = 0$.$7 - 6a = 0$.$6a = 7$.Therefore,$a = \frac{7}{6}$.