The distance of the point (1, 2, 0) from the line $\frac{x-3}{2} = \frac{y+4}{3} = \frac{z+6}{5}$ measured parallel to the plane $x + y + z = 3$ is

If the lines $\frac{1-x}{3} = \frac{3y-6}{k} = \frac{3-z}{-2}$ and $\frac{1-x}{2k} = \frac{y-5}{3} = \frac{6-z}{5}$ are perpendicular to each other, then $k$ is equal to

The angle between the lines

$l_1: \frac{x + 1}{1} = \frac{2 - y}{2} = \frac{z - 1}{1}$ and $l_2: \frac{x - 1}{4} = \frac{2y - 4}{6} = \frac{z - 1}{2}$ is

Which of the following statements are correct?

(A) The angle between the vectors with direction ratios proportional to -3, 5, 4 and 4, 5, 3 is $\dfrac{\pi}{3}$

(B) The direction cosines of Y-axis is 0, 1, 0.

(C) The cartesian equation of a line is $3x-6=2y+1=2z-1$, then its direction ratios are proportional to 2, 2, 3

(D) The vector equation of the line passing through the point $(2,-1,-1)$ and direction ratios 1, 2, 3 is $\vec r=(2\hat i-\hat j-\hat k)+\lambda(\hat i+2\hat j+3\hat k)$

Choose the correct answer from the options given below:

If a line makes angles $\alpha, \beta, \gamma$ with the positive directions of $x$-axis, $y$- axis and $z$-axis respectively, then $sin^2 \alpha + sin^2 \beta + sin^2 \gamma$ is equal to

If the lines $\frac{x-1}{2} = \frac{y+1}{\lambda-3} = \frac{z-2}{2}$ and $x = 0, \frac{2y-4}{-2} = \frac{z+1}{2-\lambda}$ are perpendicular, then value of $\lambda$ is

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