Let a line having direction ratios $1, -4, 2$ intersect the lines $\frac{x-7}{3}=\frac{y-1}{-1}=\frac{z+2}{1}$ and $\frac{x}{2}=\frac{y-7}{3}=\frac{z}{1}$ at the points $A$ and $B$ respectively. Then $( AB )^{2}$ is equal to

The line $L_1$ is parallel to the vector $\vec{a} = -3 \hat{i} + 2 \hat{j} + 4 \hat{k}$ and passes through the point $(7, 6, 2)$,and the line $L_2$ is parallel to the vector $\vec{b} = 2 \hat{i} + \hat{j} + 3 \hat{k}$ and passes through the point $(5, 3, 4)$. The shortest distance between the lines $L_1$ and $L_2$ is:

If lines $\frac{x - 1}{3} = \frac{y - 2}{-1} = \frac{z - \lambda}{2}$ and $\frac{x + 1}{-2} = \frac{y}{3\lambda} = \frac{2z - 7}{1}$ are coplanar,then the sum of the value$(s)$ of $\lambda$ is:

If the distance of the point $(a, 2, 5)$ from the image of the point $(1, 2, 7)$ in the line $\frac{x-1}{1} = \frac{y-1}{1} = \frac{z-2}{2}$ is $4$, then the sum of all possible values of $a$ is equal to :

The vector equation of the line passing through $P(1, 2, 3)$ and $Q(2, 3, 4)$ is

Let the foot of the perpendicular from the point $(\lambda, 2, 3)$ on the line $\frac{x-4}{1} = \frac{y-9}{2} = \frac{z-5}{1}$ be the point $(1, \mu, 2)$. Then the distance between the lines $\frac{x-1}{2} = \frac{y-2}{3} = \frac{z+4}{6}$ and $\frac{x-\lambda}{2} = \frac{y-\mu}{3} = \frac{z+5}{6}$ is equal to: