Two lines $L_1: x=5, \frac{y}{3-\alpha}=\frac{z}{-2}$ and $L_2: x=\alpha, \frac{y}{-1}=\frac{z}{2-\alpha}$ are coplanar. Then $\alpha$ can take value$(s)$.

If $A(0,3,4), B(1,5,6), C(-2,0,-2)$ are the vertices of a triangle $ABC$ and the bisector of angle $A$ meets the side $BC$ at $D$, then $AD=$

The line $l_1$ passes through the point $(2, 6, 2)$ and is perpendicular to the plane $2x + y - 2z = 10$. Then the shortest distance between the line $l_1$ and the line $\frac{x + 1}{2} = \frac{y + 4}{-3} = \frac{z}{2}$ is :

$\triangle ABC$ is formed by $A(1, 8, 4)$,$B(0, -11, 4)$ and $C(2, -3, 1)$. If $D$ is the foot of the perpendicular from $A$ to $BC$,then the coordinates of $D$ are

If $A(3,4,5)$, $B(4,6,3)$, $C(-1,2,4)$, and $D(1,0,5)$ are points such that the angle between the lines $DC$ and $AB$ is $\theta$, then $\cos \theta$ is equal to

Let $(\alpha, \beta, \gamma)$ be the coordinates of the foot of the perpendicular drawn from the point $(5, 4, 2)$ on the line $\vec{r} = (-\hat{i} + 3\hat{j} + \hat{k}) + \lambda(2\hat{i} + 3\hat{j} - \hat{k})$. Then the length of the projection of the vector $\alpha\hat{i} + \beta\hat{j} + \gamma\hat{k}$ on the vector $6\hat{i} + 2\hat{j} + 3\hat{k}$ is: