Let the shortest distance between the lines

$$L: \frac{x-5}{-2}=\frac{y-\lambda}{0}=\frac{z+\lambda}{1}, \lambda \geq 0$$ and

$$L_{1}: x+1=y-1=4-z$$ be $$2 \sqrt{6}$$. If $$(\alpha, \beta, \gamma)$$ lies on $$L$$,

then which of the following is NOT possible?

Let $$\vec{a}=2 \hat{i}+\hat{j}+\hat{k}$$, and $$\vec{b}$$ and $$\vec{c}$$ be two nonzero vectors such that $$|\vec{a}+\vec{b}+\vec{c}|=|\vec{a}+\vec{b}-\vec{c}|$$ and $$\vec{b} \cdot \vec{c}=0$$. Consider the following two statements:

(A) $$|\vec{a}+\lambda \vec{c}| \geq|\vec{a}|$$ for all $$\lambda \in \mathbb{R}$$.

(B) $$\vec{a}$$ and $$\vec{c}$$ are always parallel.

Then,

Let $$A = \left( {\matrix{ 1 & 0 & 0 \cr 0 & 4 & { - 1} \cr 0 & {12} & { - 3} \cr } } \right)$$. Then the sum of the diagonal elements of the matrix $${(A + I)^{11}}$$ is equal to :

For all $$z \in C$$ on the curve $$C_{1}:|z|=4$$, let the locus of the point $$z+\frac{1}{z}$$ be the curve $$\mathrm{C}_{2}$$. Then :