Coplanarity of Two Lines

The line, that is coplanar to the line $\frac{x+3}{-3}=\frac{y-1}{1}=\frac{z-5}{5}$, is

Given $\frac{x+3}{-3}=\frac{y-2}{2}=\frac{z-5}{5}$. Which of the following lines in options is coplanar with the given line?

Let $L_1,L_2$ and $L_3$ be lines given by

$\frac{x-1}{3}=\frac{y-2}{1}=\frac{z-3}{2}, \frac{x+1}{-1}=\frac{y-2}{2}=\frac{z-5}{5}$ and $\frac{x+3}{-3}=\frac{y-1}{1}=\frac{z-5}{5}$

Then,

The set of all non-zero real values of $k$, for which the lines $\frac{x-4}{2}=\frac{y-6}{2}=\frac{z-8}{-2k^2}$ and $\frac{x-2}{2k^2}=\frac{y-8}{4}=\frac{z-10}{2}$ are coplanar

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

If the line $\frac{x-2}{1}=\frac{y-3}{1}=\frac{z-4}{-k}$ and $\frac{x-1}{k}=\frac{y-4}{2}=\frac{z-5}{1}$ are coplanar, then $k$ can have

If the lines $\frac{\text{x}-2}{1}=\frac{\text{y}-3}{1}=\frac{\text{z}-4}{-\text{k}}$ and $\frac{\text{x}-1}{\text{k}}=\frac{\text{y}-4}{2}=\frac{\text{z}-5}{1}$ are coplanar, then $k$ can have

The vector equation of the plane passing through the points $R(2,5,-3), S(-2,-3,5)$ and $T(5,3,-3)$ is

The three lines through the origin with direction cosines $l_1, m_1,n_1;l_2, m_2,n_2;$ and $l_3, m_3,n_3$ are coplanar, if

If the lines $\overrightarrow{\mathrm{r}}=-\hat{\mathrm{i}}+\hat{\mathrm{j}}-\hat{\mathrm{k}}+\lambda (2\hat{\mathrm{i}}+\hat{\mathrm{j}}+3\hat{\mathrm{k}})$ and $\overrightarrow{\mathrm{r}}=-2\hat{\mathrm{i}}+\alpha \hat{\mathrm{j}}+\hat{\mathrm{k}}+\mu (2\hat{\mathrm{i}}+3\hat{\mathrm{j}}+4\hat{\mathrm{k}}); (\lambda , \mu \in \mathrm{R})$ are coplanar, then the value of $\alpha$, is

If a line passing through $(2,1,4)$ cuts off an intercept of minimum length between two non-coplanar lines $x-6=\frac{y}{\alpha }=-z$ and $x=\frac{y}{0}=z$, then $\alpha$ is

If $A\left(\overrightarrow{a}\right),B\left(\overrightarrow{b}\right),C\left(\overrightarrow{c}\right)$ and $D\left(\overrightarrow{d}\right)$ are four points such that $\overrightarrow{a}=-2\widehat{i}+4\widehat{j}+3\widehat{k},\overrightarrow{b}=2\widehat{i}-8\widehat{j},\overrightarrow{c}=\widehat{i}-3\widehat{j}+5\widehat{k}$ and $\overrightarrow{d}=4\widehat{i}+\widehat{j}-7\widehat{k}$ and $d_1$ is the shortest distance between the lines  $AB$ and $CD$, then which of the following is true:

If the lines $x=-1+a, y=3-\lambda a, z=1+\lambda a$ and $x=\frac{b}{2}, y=1+b, z=2-b$ are coplanar, then find the value of $\lambda$.

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

The shortest distance between the two skew lines $\overrightarrow{r}=(4\hat{i}-\hat{j})+\lambda (\hat{i}+2\hat{j}-3\hat{k}), \lambda \in R$ and $\overrightarrow{r}=(\hat{i}-\hat{j}+2\hat{k})+\mu (2\hat{i}+4\hat{j}-5\hat{k}), \mu \in R$ is _____.

Cotyledons are also called-

If the lines $\frac{x-3}{2}=\frac{y-5}{2}=\frac{z-4}{\lambda }$ and $\frac{x-2}{\lambda }=\frac{y-6}{4}=\frac{z-5}{2}$ intersect at a point $\left(\alpha ,\beta ,\gamma \right)$, then the greatest value of $\lambda$ is equal to

The lines $\frac{x-a+d}{\mathrm{\alpha }-\mathrm{\delta }}=\frac{y-a}{\mathrm{\alpha }}=\frac{z-a-d}{\mathrm{\alpha }+\mathrm{\delta }}$ and $\frac{x-b+c}{\mathrm{\beta }-\mathrm{\gamma }}=\frac{y-b}{\mathrm{\beta }}=\frac{z-b-c}{\mathrm{\beta }+\mathrm{\gamma }}$ are coplanar and then equation to the plane in which they lie, is

The equation of the plane passing through the lines $\frac{x-4}{1}=\frac{y-3}{1}=\frac{z-2}{2}$ and $\frac{x-3}{1}=\frac{y-2}{-4}=\frac{z}{5}$ is

The direction cosines of three lines passing through the origin are $l_1,m_1,n_1;\mathrm{ }{l}_2,m_2,n_2$ and $l_3,m_3,n_3$. The lines will be coplanar, if