The vectors $\bar{a}, \overline{\mathrm{~b}}$ and $\overline{\mathrm{c}}$ are such that $|\bar{a}|=2,|\overline{\mathrm{~b}}|=4,|\overline{\mathrm{c}}|=4$. If the projection of $\overline{\mathrm{b}}$ on $\bar{a}$ is equal to projection of $\overline{\mathrm{c}}$ on $\bar{a}$ and $\overline{\mathrm{b}}$ is perpendicular to $\overline{\mathrm{c}}$, then the value of $|\bar{a}+\overline{\mathrm{b}}-\overline{\mathrm{c}}|$ is

Given vectors $\bar{a}$, $\overline{\mathrm{b}}$, and $\overline{\mathrm{c}}$ with magnitudes $|\bar{a}| = 2$, $|\overline{\mathrm{b}}| = 4$, and $|\overline{\mathrm{c}}| = 4$.

The projection of $\overline{\mathrm{b}}$ on $\bar{a}$ equals that of $\overline{\mathrm{c}}$ on $\bar{a}$, so $\overline{\mathrm{b}} \cdot \bar{a} = \overline{\mathrm{c}} \cdot \bar{a}$, which implies $\bar{a} \cdot (\overline{\mathrm{b}} - \overline{\mathrm{c}}) = 0$.

Also, $\overline{\mathrm{b}} \perp \overline{\mathrm{c}}$, so $\overline{\mathrm{b}} \cdot \overline{\mathrm{c}} = 0$.

To compute $|\bar{a} + \overline{\mathrm{b}} - \overline{\mathrm{c}}|$, square and expand:

$|\bar{a} + \overline{\mathrm{b}} - \overline{\mathrm{c}}|^2 = |\bar{a}|^2 + |\overline{\mathrm{b}} - \overline{\mathrm{c}}|^2 + 2\bar{a} \cdot (\overline{\mathrm{b}} - \overline{\mathrm{c}})$.

Since $\bar{a} \cdot (\overline{\mathrm{b}} - \overline{\mathrm{c}}) = 0$, this simplifies to $|\bar{a}|^2 + |\overline{\mathrm{b}} - \overline{\mathrm{c}}|^2$.

Now, $|\overline{\mathrm{b}} - \overline{\mathrm{c}}|^2 = |\overline{\mathrm{b}}|^2 + |\overline{\mathrm{c}}|^2 - 2\overline{\mathrm{b}} \cdot \overline{\mathrm{c}}$. Given $\overline{\mathrm{b}} \cdot \overline{\mathrm{c}} = 0$, it reduces to $|\overline{\mathrm{b}}|^2 + |\overline{\mathrm{c}}|^2 = 16 + 16 = 32$.

Substitute to get $|\bar{a} + \overline{\mathrm{b}} - \overline{\mathrm{c}}|^2 = 4 + 32 = 36$, so the magnitude is $6$.

Answer: 6