Let $$S_{K}=\frac{1+2+\ldots+K}{K}$$ and $$\sum_\limits{j=1}^{n} S_{j}^{2}=\frac{n}{A}\left(B n^{2}+C n+D\right)$$, where $$A, B, C, D \in \mathbb{N}$$ and $$A$$ has least value. Then
If the equation of the plane containing the line
$$x+2 y+3 z-4=0=2 x+y-z+5$$ and perpendicular to the plane
$\vec{r}=(\hat{i}-\hat{j})+\lambda(\hat{i}+\hat{j}+\hat{k})+\mu(\hat{i}-2 \hat{j}+3 \hat{k})$
is $a x+b y+c z=4$, then $$(a-b+c)$$ is equal to :
Let $$P=\left[\begin{array}{cc}\frac{\sqrt{3}}{2} & \frac{1}{2} \\ -\frac{1}{2} & \frac{\sqrt{3}}{2}\end{array}\right], A=\left[\begin{array}{ll}1 & 1 \\ 0 & 1\end{array}\right]$$ and $$Q=P A P^{T}$$. If $$P^{T} Q^{2007} P=\left[\begin{array}{ll}a & b \\ c & d\end{array}\right]$$, then $$2 a+b-3 c-4 d$$ equal to :
If for $$z=\alpha+i \beta,|z+2|=z+4(1+i)$$, then $$\alpha+\beta$$ and $$\alpha \beta$$ are the roots of the equation :