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:

Let a line L passing through the point (1, 1, 1) be perpendicular to both the vectors $2\hat{i} + 2\hat{j} + \hat{k}$ and $\hat{i} + 2\hat{j} + 2\hat{k}$. If $P(a, b, c)$ is the foot of perpendicular from the origin on the line L, then the value of $34(a + b + c)$ is :

Let Q(a, b, c) be the image of the point P(3, 2, 1) in the line $\frac{x-1}{1} = \frac{y}{2} = \frac{z-1}{1}$. Then the distance of Q from the line $\frac{x-9}{3} = \frac{y-9}{2} = \frac{z-5}{-2}$ is