The value of 'a' so that the volume of the parallelopiped formed by $\hat i+a\hat j+\hat k, \hat j + a\hat k$ and $a\hat i + \hat k$ becomes minimum, is

$\frac{1}{\sqrt{3}}$

Let V be the volume of the parallelopiped formed by the given vectors.

Let $\vec α = \hat i +a\hat j+\hat k, \vec β=\hat j+a\hat k$ and $\vec γ = a\hat i + \hat k$. Then,

$[\vec α\,\,\vec β\,\,\vec γ]=\begin{vmatrix}1&a&1\\0&1&a\\a&0&1\end{vmatrix}=1+ a^3 -a$

$∴V=\left|[\vec α\,\,\vec β\,\,\vec γ]\right|$

Clearly, V will be minimum when $λ = [\vec α\,\,\vec β\,\,\vec γ]=1+ a^3 -a$ is minimum.

Now, $λ =1+ a^3 -a$

$⇒\frac{dλ}{da}=3a^2-1$ and $\frac{d^2λ}{da^2}=6a$

For maximum or minimum value of λ, we must have

$\frac{dλ}{da}=0⇒a=±\frac{1}{\sqrt{3}}$

Clearly, $(\frac{d^2λ}{da^2})_{a=1/\sqrt{3}}=\frac{6}{\sqrt{3}}>0$

So, $λ$ is minimum for $a =\frac{1}{\sqrt{3}}$

Hence, V is minimum for $a = 1/\sqrt{3}$.