$$\int_\limits1^2 \frac{\mathrm{d} x}{\left(x^2-2 x+4\right)^{\frac{3}{2}}}=\frac{\mathrm{k}}{\mathrm{k}+5} \text {, then } \mathrm{k} \text { has the value }$$

The sides of a rectangle are given by the equations $$x=-2, x=4, y=-2$$ and $$y=5$$

Then the equation of the circle, whose centre is the point of intersection of the diagonals, lying within the rectangle and touching only two opposite sides, is

$$\overline{\mathrm{a}}=\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}}, \overline{\mathrm{b}}=\hat{\mathrm{j}}-\hat{\mathrm{k}}$$, then vector $$\overline{\mathrm{r}}$$ satisfying $$\overline{\mathrm{a}} \times \overline{\mathrm{r}}=\overline{\mathrm{b}}$$ and $$\overline{\mathrm{a}} \cdot \overline{\mathrm{r}}=3$$ is

The magnitude of the projection of the vector $$2 \hat{i}+\hat{j}+\hat{k}$$ on the vector perpendicular to the plane containing the vectors $$\hat{i}+\hat{j}+\hat{k}$$ and $$\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}}$$ is