Let $\bar{a}=\hat{i}+\hat{\mathrm{j}}+\hat{\mathrm{k}}, \overline{\mathrm{b}}$ and $\overline{\mathrm{c}}=\hat{\mathrm{j}}-\hat{\mathrm{k}}$ be three vectors such that $\bar{a} \times \overline{\mathrm{b}}=\overline{\mathrm{c}}$ and $\bar{a} \cdot \bar{c}=0$. If the length of projection vector of the vector $\overline{\mathrm{b}}$ on the vector $\bar{a} \times \overline{\mathrm{c}}$ is $l$, then the value of $3 l^2$ is

Given vectors:

$\bar{a} = \hat{i} + \hat{j} + \hat{k}$

$\bar{c} = \hat{j} - \hat{k}$

The condition $\bar{a} \cdot \bar{c} = 0$ is satisfied since $\bar{a}$ and $\bar{c}$ are orthogonal.

Let $\bar{d} = \bar{a} \times \bar{c}$. Using the determinant form:

$\bar{d} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 1 \\ 0 & 1 & -1 \end{vmatrix} = -2\hat{i} + \hat{j} + \hat{k}$

The magnitude is $|\bar{d}| = \sqrt{(-2)^2 + 1^2 + 1^2} = \sqrt{6}$.

The projection length of $\bar{b}$ on $\bar{d}$ is $l = \frac{|\bar{b} \cdot \bar{d}|}{|\bar{d}|}$.

Using the scalar triple product property:

$\bar{b} \cdot (\bar{a} \times \bar{c}) = -[\bar{a}\ \bar{b}\ \bar{c}] = -(\bar{a} \times \bar{b}) \cdot \bar{c}$

Given $\bar{a} \times \bar{b} = \bar{c}$, this becomes $-\bar{c} \cdot \bar{c} = -|\bar{c}|^2$.

Since $|\bar{c}|^2 = 1^2 + (-1)^2 = 2$, we have $\bar{b} \cdot \bar{d} = -2$.

The projection length is $l = \frac{2}{\sqrt{6}}$ and $3l^2 = 3 \cdot \frac{4}{6} = 2$.

The final result is $2$.