A line with direction ratios $1,-1,2$ intersects the lines $\frac{x}{2}=\frac{y}{3}=\frac{z+1}{3}$ and $\frac{x+1}{-1}=\frac{y-2}{1}=\frac{z}{4}$ at the points P and Q , respectively. If the length of the line segment PQ is $\alpha$, then $225 \alpha^2$ 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 the point A be the foot of perpendicular drawn from the point P$(a, b, 0)$ on the line

$$\frac{x-1}{2} = \frac{y-2}{1} = \frac{z-\alpha}{3}.$$

If the midpoint of the line segment PA is $$\left(0, \frac{3}{4}, -\frac{1}{4}\right),$$ then the value of $a^2 + b^2 + \alpha^2$ is equal to :

If the point of intersection of the lines $ \frac{x+1}{3} = \frac{y+a}{5} = \frac{z+b+1}{7} $ and $ \frac{x-2}{1} = \frac{y-b}{4} = \frac{z-2a}{7} $ lies on xy-plane, then the value of $a+b$ is: