Let $$\vec{a}=2 \hat{i}+5 \hat{j}-\hat{k}, \vec{b}=2 \hat{i}-2 \hat{j}+2 \hat{k}$$ and $$\vec{c}$$ be three vectors such that $$(\vec{c}+\hat{i}) \times(\vec{a}+\vec{b}+\hat{i})=\vec{a} \times(\vec{c}+\hat{i})$$. If $$\vec{a} \cdot \vec{c}=-29$$, then $$\vec{c} \cdot(-2 \hat{i}+\hat{j}+\hat{k})$$ is equal to:

Let $$\mathrm{A}(-1,1)$$ and $$\mathrm{B}(2,3)$$ be two points and $$\mathrm{P}$$ be a variable point above the line $$\mathrm{AB}$$ such that the area of $$\triangle \mathrm{PAB}$$ is 10. If the locus of $$\mathrm{P}$$ is $$\mathrm{a} x+\mathrm{by}=15$$, then $$5 \mathrm{a}+2 \mathrm{~b}$$ is :

The values of $$m, n$$, for which the system of equations

$$\begin{aligned} & x+y+z=4, \\ & 2 x+5 y+5 z=17, \\ & x+2 y+\mathrm{m} z=\mathrm{n} \end{aligned}$$

has infinitely many solutions, satisfy the equation :

If $$y(\theta)=\frac{2 \cos \theta+\cos 2 \theta}{\cos 3 \theta+4 \cos 2 \theta+5 \cos \theta+2}$$, then at $$\theta=\frac{\pi}{2}, y^{\prime \prime}+y^{\prime}+y$$ is equal to :