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 $...........$.

If the line $\frac{x - 2}{3} = \frac{y + 1}{2} = \frac{z - 1}{-1}$ intersects the plane $2x + 3y - z + 13 = 0$ at a point $P$ and the plane $3x + y + 4z = 16$ at a point $Q$,then $PQ$ is equal to

Show that the points $(\hat{i}-\hat{j}+3 \hat{k})$ and $3(\hat{i}+\hat{j}+\hat{k})$ are equidistant from the plane $\vec{r} \cdot(5 \hat{i}+2 \hat{j}-7 \hat{k})+9=0$ and lie on opposite sides of it.

Three lines $L_1: \overrightarrow{r} = \lambda \hat{i}, \lambda \in R$,$L_2: \overrightarrow{r} = \hat{k} + \mu \hat{j}, \mu \in R$,and $L_3: \overrightarrow{r} = \hat{i} + \hat{j} + v\hat{k}, v \in R$ are given. For which point$(s)$ $Q$ on $L_2$ can we find a point $P$ on $L_1$ and a point $R$ on $L_3$ such that $P, Q,$ and $R$ are collinear?

The line of intersection of the planes $x + 2y = 0$ and $y - 3z + 3 = 0$ is

Let $\lambda_1, \lambda_2$ be the values of $\lambda$ for which the points $\left(\frac{5}{2}, 1, \lambda\right)$ and $(-2, 0, 1)$ are at equal distance from the plane $2x + 3y - 6z + 7 = 0$. If $\lambda_1 > \lambda_2$,then the distance of the point $(\lambda_1 - \lambda_2, \lambda_2, \lambda_1)$ from the line $\frac{x - 5}{1} = \frac{y - 1}{2} = \frac{z + 7}{2}$ is: