Consider a line $L$ passing through the points $P(1, 2, 1)$ and $Q(2, 1, -1)$. If the mirror image of the point $A(2, 2, 2)$ in the line $L$ is $(\alpha, \beta, \gamma)$,then $\alpha + \beta + 6\gamma$ is equal to:

The square of the distance of the point $(-2, -8, 6)$ from the line $\frac{x-1}{1} = \frac{y-1}{2} = \frac{z}{-1}$ along the line $\frac{x+5}{1} = \frac{y+5}{-1} = \frac{z}{2}$ is equal to:

Let a line passing through the point $P(4,1,0)$ intersect the line $L_1: \frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}$ at the point $A(\alpha, \beta, \gamma)$ and the line $L_2: x-6=y=-z+4$ at the point $B(a, b, c)$. Then $\left|\begin{array}{lll}1 & 0 & 1 \\ \alpha & \beta & \gamma \\ a & b & c \end{array}\right|$ is equal to

Find the point of intersection of the lines $\frac{x - 4}{5} = \frac{y - 1}{2} = \frac{z}{1}$ and $\frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{4}$.

$A(1, -2, 1)$ and $B(2, -1, 2)$ are the end points of a line segment. If $D(\alpha, \beta, \gamma)$ is the foot of the perpendicular drawn from $C(1, 2, 3)$ to $AB$,then $\alpha^2 + \beta^2 + \gamma^2 =$

The vertices $B$ and $C$ of a triangle $ABC$ lie on the line $\frac{x}{1}=\frac{1-y}{-2}=\frac{z-2}{3}$. The coordinates of $A$ and $B$ are $(1, 6, 3)$ and $(4, 9, \alpha)$ respectively, and $C$ is at a distance of $10$ units from $B$. The area (in sq. units) of $\Delta ABC$ is: