The direction ratios of the line perpendicular to the lines $\dfrac{x-5}{2} = \dfrac{y+11}{-3} = \dfrac{z+3}{1}$and $\dfrac{x-7}{1} = \dfrac{y+2}{2} = \dfrac{z-4}{-2}$ are proportional to:

If the lines $\frac{x - 5}{5\lambda + 2} = \frac{2 - y}{5} = \frac{1 - z}{-1}$ and $x = \frac{y + 1/2}{2\lambda} = \frac{z - 1}{3}$ are perpendicular, then the value of $\lambda$ is equal to

Which of the following statements are true?

(A) The vector equation of the line through the point (5, 2, -4) and parallel to the vector $3\hat{i} + 2\hat{j} - 8\hat{k}$ is $\vec{r} = (5\hat{i} + 2\hat{j} - 4\hat{k}) + \lambda(3\hat{i} + 2\hat{j} - 8\hat{k})$

(B) Vector form of the equation of line $\frac{x-5}{3} = \frac{y+4}{7} = \frac{z-6}{2}$ is $\vec{r} = (5\hat{i} - 4\hat{j} + 6\hat{k}) + \lambda(3\hat{i} + 7\hat{j} + 2\hat{k})$

(C) The direction cosines of z-axis are (1, 1,0).

(D) If a line has direction ratios 2, -1, -2, then its direction cosines are -2/3, -1/3, -2/3.

Choose the correct answer from the options given below:

The value of $\lambda$ so that the lines $\frac{1-x}{3} = \frac{7y-14}{2\lambda} = \frac{z-3}{2}$ and $\frac{7-7x}{3\lambda} = \frac{y-5}{1} = \frac{6-z}{5}$ are at right angle, is:

The foot of perpendicular from the point P (1, 2, -3) to the line $\frac{x+1}{2} = \frac{y-3}{-2} = \frac{z}{-1}$ is

Let ABCD be a parallelogram with vertices $A(2,2,a), B(1,0,0), C(1,3,a)$ and D, where $a > 0$. If the area of parallelogram is 5 square units, then the value of $a$ is:

The value of p so that the lines $\frac{x-1}{-3} = \frac{2y-2}{2p} = \frac{z-3}{2}$ and $\frac{x-1}{-3p} = \frac{y-1}{4} = \frac{6-z}{5}$ are at right angles is