The equation of the line tangential to the curve represented by y = 3x^2 - x^3 at the point (1,2) has to be determined.

For a curve y = f(x), the slope of the tangent drawn at a point where x = a is given by f'(a). A line with slope m, passing through the point `(x_1, y_1)` is `(y - y_1)/(x - x_1) = m` .

The derivative of y = 3x^2 - x^3 is:

y' = 6x - 3x^2

At the point (1,2), as x = 1, the value of the slope is 6 - 3 = 3

The equation of the tangent line is `(y - 2)/(x - 1) = 3`

y - 2 = 3x - 3

3x - y - 1 = 0

The graph of the curve and the tangent line at (1,2) is:

The required equation of the tangent is 3x - y - 1 = 0