Hint: Use triangle inequality to find the general points of parabola in parametric form to find the sum of $K_1$ and $K_2$.

Complete step-by-step answer:

First draw the diagram of the parabola equation which is horizontally parabola opening right side.

P point is given as $({K_1}a,{K_2}a)$,Q point is given as $(0,a)$ and R point is given as $(- a,0)$

When you join these points it makes a triangle, So according to triangle inequality the modulus of difference of two sides is always less than or equal to the third side.

$ \Rightarrow \left| {PR - PQ} \right| \leqslant \left| {QR} \right| $

Maximum value of $\left| {PR - PQ} \right|\; = QR$ because everything is less than or equal to QR.

$\Rightarrow \left| {PR - PQ} \right|\; = QR$, when P, Q, R are collinear

So, the equation of line joining R and Q is

$\Rightarrow y - 0 = \dfrac{{a - 0}}{{0 - \left( { - a} \right)}}\left( {x - \left( { - a} \right)} \right) $

$ \Rightarrow y = x + a $

Equation of parabola is $y^2 = 4ax $

In parametric form the coordinates of parabola is $( {a{t^2},2at}) $

Where t is a parameter and these parametric coordinates represents every point on parabola.

$ \Rightarrow$ p point is $( {a{t^2},2at}) $

Now P, Q, R is in straight line, then this p point satisfy the equation of straight line

$ \Rightarrow 2at - a{{\text{t}}^2} = a $

$ \Rightarrow {t^2} - 2t + 1 = 0 $

$\Rightarrow {\left( {t - 1} \right)^2} = 0 $

$ \Rightarrow \left( {t - 1} \right) = 0 $

$ \Rightarrow t = 1 $

So, put the value of t in P points

$ \Rightarrow {\text{P}}\left( {a,2a} \right)$, So compare this points with given points $({K_1}a,{K_2}a) $

So, on comparing $K_1 = 1, K_2 = 2 $

$ \Rightarrow$ The value of $\left( {{{\text{K}}_1} + {{\text{K}}_2}} \right) = 1 + 2 = 3 $

So, this is your answer.

NOTE: - In this type of problem always apply inequality identity, after that find out the general points of parabola in parametric form and satisfy these points in the equation which pass from this point.