$$ \bar{a}\cdot \bar{c}=0 $$

and

$$ \bar{b}\cdot \bar{c}=0 $$

Now use the given condition:

$$ |\bar{a}+\bar{b}+\bar{c}|=1 $$

Squaring both sides,

$$ |\bar{a}+\bar{b}+\bar{c}|^2=1 $$

Using the formula,

$$ |\bar{a}+\bar{b}+\bar{c}|^2 = |\bar{a}|^2+|\bar{b}|^2+|\bar{c}|^2 +2\bar{a}\cdot\bar{b} +2\bar{b}\cdot\bar{c} +2\bar{c}\cdot\bar{a} $$

Since $\bar{c}$ is perpendicular to both $\bar{a}$ and $\bar{b}$,

$$ \bar{b}\cdot\bar{c}=0 $$

and

$$ \bar{c}\cdot\bar{a}=0 $$

So,

$$ 1=|\bar{a}|^2+|\bar{b}|^2+|\bar{c}|^2+2\bar{a}\cdot\bar{b} $$

Substitute the given magnitudes:

$$ 1=\left(\frac{1}{\sqrt{3}}\right)^2+\left(\frac{1}{\sqrt{2}}\right)^2+\left(\frac{1}{\sqrt{6}}\right)^2+2\bar{a}\cdot\bar{b} $$

$$ 1=\frac{1}{3}+\frac{1}{2}+\frac{1}{6}+2\bar{a}\cdot\bar{b} $$

Now,

$$ \frac{1}{3}+\frac{1}{2}+\frac{1}{6}=1 $$

So,

$$ 1=1+2\bar{a}\cdot\bar{b} $$

Hence,

$$ \bar{a}\cdot\bar{b}=0 $$

Now,

$$ \bar{a}\cdot\bar{b}=|\bar{a}||\bar{b}|\cos\theta $$

Since $|\bar{a}|$ and $|\bar{b}|$ are non-zero,

$$ \cos\theta=0 $$

Therefore,

$$ \theta=\frac{\pi}{2} $$

So, the angle between $\bar{a}$ and $\bar{b}$ is

$$ \boxed{\frac{\pi}{2}} $$

Correct option: D.

Vector Algebra

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