Let the point $A$ be the foot of the perpendicular drawn from the point $P(a, b, 0)$ to the line $\frac{x-1}{2} = \frac{y-2}{1} = \frac{z-\alpha}{3}$. If the midpoint of the line segment $PA$ is $(0, \frac{3}{4}, -\frac{1}{4})$, then the value of $a^2+b^2+\alpha^2$ is equal to:

The equation of the straight line passing through the points $(a, b, c)$ and $(a - b, b - c, c - a)$ is:

Line $L_1$ passes through the point $(1, 2, 3)$ and is parallel to the $z$-axis. Line $L_2$ passes through the point $(\lambda, 5, 6)$ and is parallel to the $y$-axis. Let for $\lambda = \lambda_1, \lambda_2$ with $\lambda_2 < \lambda_1$,the shortest distance between the two lines be $3$. Then the square of the distance of the point $(\lambda_1, \lambda_2, 7)$ from the line $L_1$ is:

The equation of the $x$-axis is:

The shortest distance between the lines $\frac{x+1}{3}=\frac{y-2}{2}=\frac{z+1}{2}$ and $\frac{x-2}{1}=\frac{y-2}{2}=\frac{z+3}{3}$ is

If the direction cosines of two lines are given by $l+m+n=0$ and $mn-2lm-2nl=0$,then the acute angle between those lines is