Let $$\vec{a}=3 \hat{i}+\hat{j}-2 \hat{k}, \vec{b}=4 \hat{i}+\hat{j}+7 \hat{k}$$ and $$\vec{c}=\hat{i}-3 \hat{j}+4 \hat{k}$$ be three vectors. If a vectors $$\vec{p}$$ satisfies $$\vec{p} \times \vec{b}=\vec{c} \times \vec{b}$$ and $$\vec{p} \cdot \vec{a}=0$$, then $$\vec{p} \cdot(\hat{i}-\hat{j}-\hat{k})$$ is equal to

The sum of the series $$\frac{1}{1-3 \cdot 1^2+1^4}+\frac{2}{1-3 \cdot 2^2+2^4}+\frac{3}{1-3 \cdot 3^2+3^4}+\ldots$$ up to 10 -terms is

Two marbles are drawn in succession from a box containing 10 red, 30 white, 20 blue and 15 orange marbles, with replacement being made after each drawing. Then the probability, that first drawn marble is red and second drawn marble is white, is

The distance of the point $$Q(0,2,-2)$$ form the line passing through the point $$P(5,-4, 3)$$ and perpendicular to the lines $$\vec{r}=(-3 \hat{i}+2 \hat{k})+\lambda(2 \hat{i}+3 \hat{j}+5 \hat{k}), \lambda \in \mathbb{R}$$ and $$\vec{r}=(\hat{i}-2 \hat{j}+\hat{k})+\mu(-\hat{i}+3 \hat{j}+2 \hat{k}), \mu \in \mathbb{R}$$ is :