endicular to the plane containing $\bar a,\bar b,\bar c$.

The angle between $\bar a$ and $\bar b$ is $30^\circ$, so

$$ |\bar a \times \bar b|=|\bar a||\bar b|\sin 30^\circ $$

Since $\bar a$ and $\bar b$ are unit vectors,

$$ |\bar a \times \bar b|=1\cdot 1 \cdot \frac{1}{2}=\frac{1}{2} $$

Also, $\bar a \times \bar b$ is perpendicular to the plane of $\bar a,\bar b,\bar c$. So it is along $\bar d$.

Thus we can take

$$ \bar a \times \bar b=\frac{1}{2}\bar d $$

Now,

$$ (\bar a \times \bar b)\times(\bar c \times \bar d) = \frac{1}{2}\bar d \times (\bar c \times \bar d) $$

Using the vector triple product formula,

$$ \bar d \times (\bar c \times \bar d) = \bar c(\bar d\cdot \bar d)-\bar d(\bar d\cdot \bar c) $$

Since $\bar d$ is a unit vector,

$$ \bar d\cdot \bar d=1 $$

And since $\bar d$ is perpendicular to $\bar c$,

$$ \bar d\cdot \bar c=0 $$

So,

$$ \bar d \times (\bar c \times \bar d)=\bar c $$

Hence,

$$ (\bar a \times \bar b)\times(\bar c \times \bar d)=\frac{1}{2}\bar c $$

Given,

$$ (\bar a \times \bar b)\times(\bar c \times \bar d) = \frac{3}{26}\hat i-\frac{2}{13}\hat j+\frac{6}{13}\hat k $$

So,

$$ \frac{1}{2}\bar c = \frac{3}{26}\hat i-\frac{2}{13}\hat j+\frac{6}{13}\hat k $$

Multiplying by $2$,

$$ \bar c = \frac{3}{13}\hat i-\frac{4}{13}\hat j+\frac{12}{13}\hat k $$

So the correct option is

$$ \boxed{\text{Option A}} $$

Vector Algebra

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