Let $(\alpha, \beta, \gamma)$ be the co-ordinates of the foot of the perpendicular drawn from the point $(5,4,2)$ on the line $\overrightarrow{\mathrm{r}}=(-\hat{i}+3 \hat{j}+\hat{k})+\lambda(2 \hat{i}+3 \hat{j}-\hat{k})$.

Then the length of the projection of the vector $\alpha \hat{i}+\beta \hat{j}+\gamma \hat{k}$ on the vector $6 \hat{i}+2 \hat{j}+3 \hat{k}$ is :

Let $\overrightarrow{\mathrm{c}}$ and $\overrightarrow{\mathrm{d}}$ be vectors such that $|\overrightarrow{\mathrm{c}}+\overrightarrow{\mathrm{d}}|=\sqrt{29}$ and $\overrightarrow{\mathrm{c}} \times(2 \hat{i}+3 \hat{j}+4 \hat{k})=(2 \hat{i}+3 \hat{j}+4 \hat{k}) \times \overrightarrow{\mathrm{d}}$. If $\lambda_1, \lambda_2\left(\lambda_1>\lambda_2\right)$ are the possible values of $(\vec{c}+\vec{d}) \cdot(-7 \hat{i}+2 \hat{j}+3 \hat{k})$, then the equation $\mathrm{K}^2 x^2+\left(\mathrm{K}^2-5 \mathrm{~K}+\lambda_1\right) x y+\left(3 \mathrm{~K}+\frac{\lambda_2}{2}\right) y^2-8 x+12 y+\lambda_2=0$ represents a circle, for K equal to :

Let $ \vec{a} = \hat{i} + 2\hat{j} + \hat{k} $ and $ \vec{b} = 2\hat{i} + \hat{j} - \hat{k} $. Let $ \hat{c} $ be a unit vector in the plane of the vectors $ \vec{a} $ and $ \vec{b} $ and be perpendicular to $ \vec{a} $. Then such a vector $ \hat{c} $ is:

Let $ \vec{a} $ and $ \vec{b} $ be the vectors of the same magnitude such that

$ \frac{|\vec{a} + \vec{b}| + |\vec{a} - \vec{b}|}{|\vec{a} + \vec{b}| - |\vec{a} - \vec{b}|} = \sqrt{2} + 1. $ Then $ \frac{|\vec{a} + \vec{b}|^2}{|\vec{a}|^2} $ is :