If the locus of mid point of any normal chord of the parabola $y^2 = 4x$ is $x - \lambda = \dfrac{\mu}{y^2} + \dfrac{y^2}{\nu}$, where $\lambda, \mu, \nu \in \mathbb{N}$, then $(\lambda + \mu + \nu)$ equals to

Let the midpoint of the normal chord be $(h, k)$.

The equation of the chord with midpoint $(h, k)$ is given by $T = S_1$:

$yk - 2(x + h) = k^2 - 4h$

$2x - ky + k^2 - 2h = 0$

The slope of this chord is $m = \dfrac{2}{k}$.

The equation of the normal to the parabola $y^2 = 4x$ (where $a = 1$) having slope $m$ is:

$y = mx - 2m - m^3$

$mx - y - 2m - m^3 = 0$

Since both equations represent the same line, their coefficients must be proportional:

$\dfrac{2}{m} = \dfrac{-k}{-1} = \dfrac{k^2 - 2h}{-2m - m^3}$

From the first two parts, we get $m = \dfrac{2}{k}$.

From the second and third parts, we get:

$k = \dfrac{k^2 - 2h}{-2m - m^3}$

$k^2 - 2h = -2mk - m^3k$

Substituting $m = \dfrac{2}{k}$:

$k^2 - 2h = -2\left(\dfrac{2}{k}\right)k - \left(\dfrac{2}{k}\right)^3k$

$k^2 - 2h = -4 - \dfrac{8}{k^2}$

$2h = k^2 + 4 + \dfrac{8}{k^2}$

$h = \dfrac{k^2}{2} + 2 + \dfrac{4}{k^2}$

Replacing $(h, k)$ with $(x, y)$, the locus is:

$x - 2 = \dfrac{4}{y^2} + \dfrac{y^2}{2}$

Comparing this with the given equation $x - \lambda = \dfrac{\mu}{y^2} + \dfrac{y^2}{\nu}$, we get:

$\lambda = 2$, $\mu = 4$, $\nu = 2$

Therefore, $\lambda + \mu + \nu = 2 + 4 + 2 = 8$.

Answer: $8$