The foot of the perpendicular from a point on the circle $x^{2} + y^{2} = 1, z = 0$ to the plane $2x + 3y + z = 6$ lies on which one of the following curves?

Let the coordinates of one vertex of $\triangle ABC$ be $A(0, 2, \alpha)$ and the other two vertices lie on the line $\frac{x+\alpha}{5} = \frac{y-1}{2} = \frac{z+4}{3}$. For $\alpha \in \mathbb{Z}$,if the area of $\triangle ABC$ is $21$ sq. units and the line segment $BC$ has length $2\sqrt{21}$ units,then $\alpha^2$ is equal to $...........$.

$A$ line with positive direction cosines passes through the point $P(2, -1, 2)$ and makes equal angles with the coordinate axes. If the line meets the plane $2x + y + z = 9$ at point $Q,$ then the length $PQ$ equals

Let $S$ be the reflection of a point $Q$ with respect to the plane given by $\vec{r} = -(t+p) \hat{i} + \hat{j} + (1+p) \hat{k}$,where $t, p$ are real parameters and $\hat{i}, \hat{j}, \hat{k}$ are the unit vectors along the three positive coordinate axes. If the position vectors of $Q$ and $S$ are $10 \hat{i} + 15 \hat{j} + 20 \hat{k}$ and $\alpha \hat{i} + \beta \hat{j} + \gamma \hat{k}$ respectively,then which of the following is/are $TRUE$?
$(A)$ $3(\alpha+\beta) = -101$
$(B)$ $3(\beta+\gamma) = -71$
$(C)$ $3(\gamma+\alpha) = -86$
$(D)$ $3(\alpha+\beta+\gamma) = -121$

$A$ line $l$ passes through the origin and is perpendicular to the lines $l_1 = (3 + t)\hat{i} + (-1 + 2t)\hat{j} + (4 + 2t)\hat{k}$ and $l_2 = (3 + 2s)\hat{i} + (3 + 2s)\hat{j} + (2 + s)\hat{k}$.
Statement $1$: Line $l$ and $l_2$ are coplanar lines.
Statement $2$: Line $l$ and $l_2$ are intersecting lines.

The image of the point $(5, 2, 6)$ with respect to the plane $x + y + z = 9$ is