Condition for Coplanarity of Four Points

Q. The lines $\frac{x-2}{1}=\frac{y-3}{1}=\frac{z-4}{-k}$ and $\frac{x-1}{k}=\frac{y-4}{2}=\frac{z-5}{1}$ are coplanar if

1. $k=1\text{or}-1$
2. $k=0\text{or}-3$
3. $k=3\text{or}-3$
4. $k=0\text{or}-1$

Q. $\begin{array}{cccc}& \text{List I}& & \text{List II}\\ \text{(I)}& \text{The value of}\underset{n\to \infty }{lim}\frac{\sqrt[3]{3n^2}\cdot \sin (n!)}{n+1}\text{is}& \text{(P)}& -1\\ \text{(II)}& \text{If}x\in [0,2\pi ]\text{and}{\log }_2\tan x+{\log }_2\tan 2x=0,& & \\ & \text{then the number of solutions is}& \text{(Q)}& 0\\ \text{(III)}& \text{An unbaised dice is thrown and the number}& & \\ & {\text{appear is put in the place of}}^{\prime }{p}^{\prime }\text{in equation}& & \\ & x^2+px+2=0.\text{If the probability of the}& & \\ & \text{equation having real roots is}\frac{a}{b},(a,b\in N)& & \\ & \text{then the least possible value of}(a+b)\text{is}& \left(R\right)& 1\\ \text{(IV)}& \text{Three lines through origin having direction}& & \\ & \text{ratios}(1,a,a^2);(1,b,b^2);(1,c,c^2)\text{are non-}& & \\ & \text{coplanar. But the lines with direction ratios}& & \\ & (a,a^2,1+a^3);(b,b^2,1+b^3);(c,c^2,1+c^3)\text{are}& & \\ & \text{coplanar, then the value of}{}^{\prime }ab{c}^{\prime }\text{is}& \left(S\right)& 2\\ & & \left(T\right)& 4\\ & & \left(U\right)& 5\end{array}$

Which of the following option is CORRECT ?

1. $\left(I\right)\to \left(R\right)$
2. $\left(II\right)\to \left(U\right)$
3. $\left(III\right)\to \left(T\right)$
4. $\left(IV\right)\to \left(P\right)$

Q.

The points A(4, 5, 1), B(0, -1, -1), C(3, 9, 4) and D(-4, 4, 4) are
[Kurukshetra CEE 2002]

1. Collinear

2. Coplanar

3. Non- coplanar

4. Non- Collinear and non-coplanar

Q. The three lines drawn from O with direction ratios, (1, –1, k), (2, –3, 0) and (1, 0, 3) are coplanar. Then k =

1. 1
2. \N
3. no such k exists
4. 7

Q. Let $S$ be the set of all real values of $\lambda$ such that plane passing through the points $(-{\lambda }^2,1,1),(1,-{\lambda }^2,1)$ and $(1,1,-{\lambda }^2)$ also passes through the point $(-1,-1,1)$. Then $S$ is equal to :

1. $\left\{\sqrt{3}\right\}$
2. $\{1,-1\}$
3. $\{3,-3\}$
4. $\{\sqrt{3},-\sqrt{3}\}$

Q. The lines $\frac{x-2}{1}=\frac{y-3}{1}=\frac{z-4}{-k}$ and $\frac{x-1}{k}=\frac{y-4}{2}=\frac{z-5}{1}$ are coplanar if

1. $k=1\text{or}-1$
2. $k=0\text{or}-3$
3. $k=3\text{or}-3$
4. $k=0\text{or}-1$