If  $\dpi{100} \vec{a},\vec{b},\vec{c}$ are non-­coplanar vectors and $\dpi{100} \lambda$ is a real number, then the vectors $\dpi{100} \vec{a}+2\vec{b}+3\vec{c},\; \lambda\vec{b}+4\vec{c}\; \; and\; \; (2\lambda -1)\vec{c}$ are non-­coplanar for Option 1) all except two values of $\lambda$ Option 2) all except one values of $\lambda$ Option 3) all values of  $\lambda$ Option 4) no value of   $\lambda$

As we learnt in

Coplanar vectors -

$\left [ \vec{a}\;\vec{b}\;\vec{c} \right ]=0$

- wherein

$\vec{a}$$\vec{b}$ and $\vec{c}$ are three vectors.

Condition for given three vectors to be

Coplanar is $\begin{vmatrix} 1 & 2 &3 \\ 0& \lambda &4 \\ 0 & 0 & 2\lambda-1 \end{vmatrix}=0$

$\Rightarrow \lambda(2\lambda -1)=0$

$\Rightarrow \lambda =0, \lambda=\frac{1}{2}$

They are non coplanar for all real $\lambda$ except $\lambda=0,\frac{1}{2}$

Option 1)

all except two values of $\lambda$

Correct

Option 2)

all except one values of $\lambda$

Incorrect

Option 3)

all values of  $\lambda$

Incorrect

Option 4)

no value of   $\lambda$

Incorrect

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