Statement- $\left(S_1\right)$: The equation $(x–a) (x – c)+\lambda (x–b) (x–d)=0$ where $a<b<c<d$     
                                       has non real roots if  $\lambda >0.$
Statement- $\left(S_2\right)$: The range of $f\left(x\right)=(K – 3)$ $x^2-2Kx+3K-6$ is [0, $\infty$)(where $x\in R$) then $K=6,$ $\frac{3}{2}$ 

Explanation:
$S_1$ :R is obviously true. To test the assertion let
$f \left(x\right)=(x–a) (x–c)+\lambda (x–b) (x–d)= 0$ then 
$f \left(a\right) = \lambda (a – b) (a – d)>0$
$f \left(c\right) = \lambda (c – b) (c – d)>0$
If $\lambda > 0$ then f(a) > 0, f(c) < 0
$\Rightarrow$  there is a root between a  and c.
Thus assertion A is false.
$S_2:D=0$

$$\begin{gathered} b^2=4ac \\ \cancel{4}{K}^2=\cancel{4}(K-3)(3K-6) \\ {K}^2=3K^2-15K+18 \\ 2K^2-15K+18=0 \\ 2K^2-12K-3K+18=0 \\ 2K(K-6)-3(K-6)=0 \\ K=6,\frac{3}{2} \end{gathered}$$