Solution:
Given that \( \vec{a}=\lambda\hat i +\hat j -\hat k \)
\( \vec{b}=3\hat i -\hat j + 2\hat k \)
And \( \vec{c} \) be a vector s.t
\( (\vec{a}+\vec{b}+\vec{c})\times\vec{c}=0 \quad ....(1) \)
\( \vec{a}\cdot\vec{c}=-17 \quad ....(2) \)
\( \vec{b}\cdot\vec{c}=-20 \quad ....(3) \)
Now, from \( (1) \) we can say that \( \vec{c} \) is parallel to \( (\vec{a}+\vec{b}+\vec{c}) \) vector
So, we can assume \( \vec{c}=k(\vec{a}+\vec{b}+\vec{c}) \Rightarrow (1-k)\vec{c}=k(\vec{a}+\vec{b}) \)
\( \Rightarrow \vec{c}=k'(\vec{a}+\vec{b}) \quad ....(4) \)
From \( (2) \) and \( (4) \), we have
\( \vec{a}\cdot k'(\vec{a}+\vec{b})=-17 \)
\( k'|\vec{a}|^{2}+k'(\vec{a}\cdot\vec{b})=-17 \)
\( k'(\lambda^{2}+2+3\lambda-3)=-17 \Rightarrow\quad k'=\dfrac{-17}{\lambda^{2}+3\lambda-1} \quad ....(5) \)
From \( (3) \) and \( (4) \) we have,
\( \vec{b}\cdot k'(\vec{a}+\vec{b})=-20 \Rightarrow k'(\vec{b}\cdot\vec{a}+|\vec{b}|^{2})=-20 \)
\( k'(3\lambda-3+14)=-20 \)
\( k'=\dfrac{-20}{11+3\lambda} \quad ....(6) \)
\( \therefore \) from \( (5) \) and \( (6) \) we get
\( \dfrac{-17}{\lambda^{2}+3\lambda-1}=\dfrac{-20}{11+3\lambda} \)
On solving, we have \( \lambda=3 \Rightarrow\quad k'=-1 \ \quad\therefore\ \vec{c}=-1(\vec{a}+\vec{b}) \)
\( =-\vec{a}-\vec{b} \Rightarrow \vec{c}=-6\hat i-\hat{k} \)
Therefore
\( \vec{c}\times(3\hat i +\hat j +\hat k) = (-6\hat i - \hat{k}) \times (3\hat i +\hat j +\hat k) \)
\( =\hat i - 3\hat j - 6\hat k \ \Rightarrow \ \left|\vec{c}\times(3\hat i +\hat j +\hat k)\right|^{2} = \left(\sqrt{1^{2}+(-3)^{2}+(-6)^{2}}\right)^{2} = 46 \)

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