Let the parabola $y = x^2 + px + q$ passing through the point $(1, -1)$ be such that the distance between its vertex and the $x$-axis is minimum. Then the value of $p^2 + q^2$ is :

Let $P = \{ \theta \in [0, 4\pi] : \tan^2 \theta \neq 1 \}$ and $S = \{ a \in \mathbb{Z} : 2(\cos^8 \theta - \sin^8 \theta) \sec 2 \theta = a^2, \theta \in P \}$. Then $n(S)$ is:

Let the vectors $\vec{a} = -\hat{i} + \hat{j} + 3\hat{k}$ and $\vec{b} = \hat{i} + 3\hat{j} + \hat{k}$. For some $\lambda, \mu \in \mathbb{R}$, let $\vec{c} = \lambda \vec{a} + \mu \vec{b}$.

If $\vec{c} \cdot (3\hat{i} - 6\hat{j} + 2\hat{k}) = 10$ and $\vec{c} \cdot (\hat{i} + \hat{j} + \hat{k}) = -2$, then $|\vec{c}|^2$ is equal to :

Let the point A be the foot of perpendicular drawn from the point P$(a, b, 0)$ on the line

$$\frac{x-1}{2} = \frac{y-2}{1} = \frac{z-\alpha}{3}.$$

If the midpoint of the line segment PA is $$\left(0, \frac{3}{4}, -\frac{1}{4}\right),$$ then the value of $a^2 + b^2 + \alpha^2$ is equal to :