Cartesian equation of plane passing through the points (2, -4, 5) and perpendicular to the line with direction ratios (3, -1, 2) is :

The reflection of the point $(\alpha, \beta, \gamma)$ in the xz plane is

If angle between the lines $\dfrac{x-1}{\propto} = \dfrac{y+2}{-5} = \dfrac{z+\frac{7}{3}}{\beta}$ and $x=z, y=0$ is $\dfrac{\pi}{4}$, then

Consider two lines $l_1$ and $l_2$ with cartesian equations $\frac{x}{2} = \frac{1-y}{-2} = \frac{z}{1}$ and $\frac{2x-5}{16} = \frac{y-2}{-1} = \frac{z-5}{4}$ respectively. Which of the following is/are true?

(A) Direction ratio of $l_1$ are 2, 2, 1

(B) Direction cosines of $l_1$ are $\frac{2}{3}, \frac{-2}{3}, \frac{1}{3}$

(C) Direction ratio of $l_2$ are 16, -1, 4

(D) Angle between $l_1$ and $l_2$ is $\cos^{-1}\left(\frac{38}{3\sqrt{273}}\right)$

Choose the correct answer from the options given below:

If the shortest distance between the lines $\vec r=(\hat i+2\hat j+3\hat k)+\lambda(2\hat i+3\hat j+4\hat k)$ and $\vec r=(a\hat i+4\hat j+5\hat k)+\mu(4\hat i+6\hat j+8\hat k)$, $a \in N$, where $\lambda$ and $\mu$ are parameters, is $\sqrt{\dfrac{5}{29}}$, then $a$ 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

If the straight lines $x = 1 + s$, $y = -3 - \lambda s$, $z = 1 + \lambda s$ and $x = \frac{t}{2}$, $y = 1 + t$, $z = 2 - t$ with parameters $s$ and $t$ respectively, are coplanar, then $\lambda$ is equal to :