The parabola  $y=x^2+k_{1}x+k_2$  and  $y=x\left(k_3-x\right)$  touch each other at the point $\left(1,0\right)$ then

$y=x^2+k_{1}x+k_2$  and $y=x\left(k_3-x\right)$

As both parabolas touch each other at $\left(1,0\right)$
Point $\left(1,0\right)$ lie son both the parabolas

$0=1+k_1+k_2\Rightarrow k_1+k_2=-1$             …(1)

$0=1\left(k_3-1\right)\Rightarrow k_3=1$  …(2)

As both curves touch each other at $\left(1,0\right)$

They will have a common tangent at $\left(1,0\right)$

Slope of both curves at $\left(1,0\right)$ should be equal hence we have $2x+1\underset{\left(1,0\right)}{\left|=\right.}{k}_3-2x\underset{\left(1,0\right)}{\left| \right.}$

$k_1+2=k_3-2$  as   $k_3=1$

$k_1+2=-1\Rightarrow k_1=-3$

$k_2=2$

$\therefore k_1=-3,k_2=2,k_3=1$