Thanks Isomorphism & Danshader
$\displaystyle y = x(x - 1)(x - 2)$

$\displaystyle y = (x^2 - x)(x - 2)$

$\displaystyle y = x^3 - 2x^2 + -x^2 + 2x$

$\displaystyle y = x^3 - 3x^2 + 2x$

$\displaystyle \frac{dy}{dx} = 3x^{3-1} - 3 \times 2x^{2-1} + 2$

$\displaystyle \frac{dy}{dx} = 3x^2 - 6x + 2$

$\displaystyle \frac{dy}{dx}_A | x = 1$

███ $\displaystyle \frac{dy}{dx}_A = 3(1)^2 - 6(1) + 2 = -1$

$\displaystyle \frac{dy}{dx}_B | x = 2$

███ $\displaystyle \frac{dy}{dx}_B = 3(2)^2 - 6(2) + 2 = -4$

Equation of tangent to the curve at the point $\displaystyle A(1,0)$

$\displaystyle y - y_1 = m(x - x_1)$

$\displaystyle y - 0 = -1(x - 1)$

$\displaystyle y = -x + 1$

Equation of tangent to the curve at the point $\displaystyle B(2,0)$

$\displaystyle y - y_1 = m(x - x_1)$

$\displaystyle y - 0 = -4(x - 2)$

$\displaystyle y = -4x + 8$

$\displaystyle y + x = 1 \longrightarrow$

equation 1
$\displaystyle y + 4x = 8 \longrightarrow$

equation 2
$\displaystyle -3x = -7$

$\displaystyle x = \frac{7}{3}$

Substitute $\displaystyle x$ in equation $\displaystyle 1$
$\displaystyle y + x = 1$

$\displaystyle y + \frac{7}{3} = 1$

$\displaystyle y = 1 - \frac{7}{3}$

$\displaystyle y = -\frac{4}{3}$

$\displaystyle C\left(\frac{7}{3},-\frac{4}{3}\right)$

Is it correct?