vec a and vec c are unit collinear vectors and |vec b|= 6, then vec b…

We have, $\vec b-3\vec c=λ\vec a$

Taking scalar product with $\vec c$, we have

$(\vec b-3\vec c). \vec c =λ(\vec a.\vec c)$

$⇒\vec b.\vec c-3(\vec c.\vec c)=λ(\vec a.\vec c)$

$⇒\vec b.\vec c-3=λ$  [$∵|\vec a|=|\vec c|=1$ and $\vec a$ and $\vec c$ are collinear vectors]

$⇒\vec b.\vec c=3+λ$

Again,

$\vec b-3\vec c=λ\vec a$

$⇒|\vec b-3\vec c|=|λ\vec a|$

$⇒|\vec b-3\vec c|^2=λ^2|\vec a|^2$

$⇒|\vec b|^2+9|\vec c|^2-6(\vec b.\vec c)=λ^2|\vec a|^2$

$⇒36+9-6(3+λ)=λ^2$  [Using (i)]

$⇒27-6λ=λ^2⇒λ^2+6λ-27=0⇒λ=-9,3$