If the image of the point $P(1, 2, a)$ in the line $\frac{x-6}{3}=\frac{y-7}{2}=\frac{7-z}{2}$ is $Q(5, b, c)$,then $a^{2}+b^{2}+c^{2}$ is equal to

The equation of the line equally inclined with the positive coordinate axes and passing through $(-3, 2, -5)$ is

The shortest distance between the lines $\overline{r} = (4\hat{i} - \hat{j}) + \lambda(\hat{i} + 2\hat{j} - 3\hat{k})$ and $\overline{r} = (\hat{i} - \hat{j} + 2\hat{k}) + \mu(2\hat{i} + 4\hat{j} - 5\hat{k})$ is:

Let the point $(-1, \alpha, \beta)$ lie on the line of the shortest distance between the lines $\frac{x+2}{-3}=\frac{y-2}{4}=\frac{z-5}{2}$ and $\frac{x+2}{-1}=\frac{y+6}{2}=\frac{z-1}{0}$. Then $(\alpha-\beta)^2$ is equal to ....................

Let $S$ be the set of all values of $\lambda$,for which the shortest distance between the lines $\frac{x-\lambda}{0}=\frac{y-3}{4}=\frac{z+6}{1}$ and $\frac{x+\lambda}{3}=\frac{y}{-4}=\frac{z-6}{0}$ is $13$. Then $8\left|\sum_{\lambda \in S} \lambda\right|$ is equal to

Find the angle between the lines $\frac{x - 2}{3} = \frac{y + 1}{-2}; z = 2$ and $\frac{x - 1}{1} = \frac{2y + 3}{3} = \frac{z + 5}{2}$.