Let the volume of a parallelepiped whose coterminous edges are given by $$\vec{u} = \hat{i} + \hat{j} + \lambda\hat{k}$$, $$\vec{v} = \hat{i} + \hat{j} + 3\hat{k}$$ and $$\vec{w} = 2\hat{i} + \hat{j} + \hat{k}$$ be 1 cu. unit. If $$\theta$$ be the angle between the edges $$\vec{u}$$ and $$\vec{w}$$, then the value of $$\cos\theta$$ can be

We recall that the volume of a parallelepiped whose coterminous edges are the vectors $$\vec u,\;\vec v,\;\vec w$$ is given by the absolute value of the scalar triple product:

$$V=\left|\vec u\cdot(\vec v\times\vec w)\right|.$$

The problem states that this volume equals $$1$$. First we compute the cross-product $$\vec v\times\vec w$$, writing it as the determinant of a $$3\times3$$ matrix:

$$\vec v\times\vec w=\begin{vmatrix} \hat{i} & \hat{j} & \hat{k}\\[2pt] 1 & 1 & 3\\[2pt] 2 & 1 & 1 \end{vmatrix}.$$

Expanding term by term, we have

$$ \vec v\times\vec w =\hat{i}(1\cdot1-3\cdot1)-\hat{j}(1\cdot1-3\cdot2)+\hat{k}(1\cdot1-1\cdot2). $$

Simplifying each bracket:

$$ 1\cdot1-3\cdot1 = -2,\qquad 1\cdot1-3\cdot2 = -5,\qquad 1\cdot1-1\cdot2 = -1. $$

So

$$ \vec v\times\vec w=-2\hat{i}+5\hat{j}-1\hat{k}. $$

Now we take the dot product with $$\vec u=\hat{i}+\hat{j}+\lambda\hat{k}$$:

$$ \vec u\cdot(\vec v\times\vec w) =(1)(-2)+(1)(5)+\lambda(-1) =-2+5-\lambda =3-\lambda. $$

Because the given volume is $$1$$, we impose

$$ |3-\lambda|=1. $$

This single equation yields two possible values of $$\lambda$$:

$$ 3-\lambda=1 \quad\Longrightarrow\quad \lambda=2, $$

or

$$ 3-\lambda=-1 \quad\Longrightarrow\quad \lambda=4. $$

Next we must find the cosine of the angle $$\theta$$ between the edges $$\vec u$$ and $$\vec w$$. Using the definition of the dot product, we state the formula:

$$ \cos\theta=\dfrac{\vec u\cdot\vec w}{\|\vec u\|\,\|\vec w\|}. $$

The vector $$\vec w$$ is fixed:

$$ \vec w = 2\hat{i}+\hat{j}+\hat{k}. $$

First we compute its magnitude:

$$ \|\vec w\| =\sqrt{2^{2}+1^{2}+1^{2}} =\sqrt{4+1+1} =\sqrt6. $$

Now we treat the two admissible values of $$\lambda$$ separately.

(i) For $$\lambda=2$$

We have $$\vec u=\hat{i}+\hat{j}+2\hat{k}$$. The dot product is

$$ \vec u\cdot\vec w =(1)(2)+(1)(1)+2(1) =2+1+2 =5. $$

The magnitude of $$\vec u$$ is

$$ \|\vec u\| =\sqrt{1^{2}+1^{2}+2^{2}} =\sqrt{1+1+4} =\sqrt6. $$

Therefore

$$ \cos\theta =\dfrac{5}{\sqrt6\,\sqrt6} =\dfrac{5}{6}. $$

This value is not present among the given options.

(ii) For $$\lambda=4$$

Now $$\vec u=\hat{i}+\hat{j}+4\hat{k}$$. The dot product is

$$ \vec u\cdot\vec w =(1)(2)+(1)(1)+4(1) =2+1+4 =7. $$

The magnitude of $$\vec u$$ becomes

$$ \|\vec u\| =\sqrt{1^{2}+1^{2}+4^{2}} =\sqrt{1+1+16} =\sqrt{18} =3\sqrt2. $$

So the cosine is

$$ \cos\theta =\frac{7}{(3\sqrt2)(\sqrt6)} =\frac{7}{3\sqrt{12}} =\frac{7}{3\cdot2\sqrt3} =\frac{7}{6\sqrt3}. $$

This value matches Option 2.

Hence, the correct answer is Option 2.