If a, b, c are non-coplanar vectors and lambd | vedclass

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(C) Since $a, b, c$ are non-coplanar vectors,their scalar triple product $[a, b, c] 
eq 0$.The vectors $a + 2b + 3c, \lambda b + 4c$,and $(2\lambda -

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All except two values of $\lambda$

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(C) Since $a, b, c$ are non-coplanar vectors,their scalar triple product $[a, b, c] 
eq 0$.The vectors $a + 2b + 3c, \lambda b + 4c$,and $(2\lambda - 1)c$ are non-coplanar if and only if their scalar triple product is non-zero:$[(a + 2b + 3c), (\lambda b + 4c), (2\lambda - 1)c] 
eq 0$.Using the properties of the scalar triple product,we can write this as:$(a + 2b + 3c) \cdot [(\lambda b + 4c) 	imes (2\lambda - 1)c] 
eq 0$$(a + 2b + 3c) \cdot [\lambda(2\lambda - 1)(b 	imes c)] 
eq 0$Since $a \cdot (b 	imes c) = [a, b, c]$,$b \cdot (b 	imes c) = 0$,and $c \cdot (b 	imes c) = 0$,the expression simplifies to:$\lambda(2\lambda - 1)[a, b, c] 
eq 0$.Given $[a, b, c] 
eq 0$,the condition for non-coplanarity is $\lambda(2\lambda - 1) 
eq 0$.This implies $\lambda 
eq 0$ and $\lambda 
eq \frac{1}{2}$.Thus,the vectors are non-coplanar for all values of $\lambda$ except $\lambda = 0$ and $\lambda = \frac{1}{2}$.

If $a, b, c$ are non-coplanar vectors and $\lambda$ is a real number,then the vectors $a + 2b + 3c, \lambda b + 4c$ and $(2\lambda - 1)c$ are non-coplanar for

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