If \(2y=x+3\) is a tangent line to the curve \(y^{3}=ax^{2}+b\) at \((1,2)\) then \(3b-5a\) is equal to

\(0\) \(2\) \(3\) \(4\)

\(0\)

\(\begin{aligned}\left(\frac{dy}{dx}\right)_{(1,2)}&=\frac{a}{6}\\ \text{Thus equation }:(y-2)&=\frac{a}{6}(x-1)\\ ax-6y&=-a+12 .....(1),\\ x-2y+3&=0......(2)\end{aligned}\hspace{5cm}\) (1) and (2) are same, So, \(a=3, b=5\)