Let $$\vec{a} = \hat{i} + \hat{j} + \hat{k}$$, $$\vec{b}$$ and $$\vec{c} = \hat{j} - \hat{k}$$ be three vectors such that $$\vec{a} \times \vec{b} = \vec{c}$$ and $$\vec{a} \cdot \vec{b} = 1$$. If the length of projection vector of the vector $$\vec{b}$$ on the vector $$\vec{a} \times \vec{c}$$ is $$l$$, then the value of $$3l^2$$ is equal to _________.

We have the given vectors

$$\vec a = \hat i + \hat j + \hat k, \qquad \vec c = \hat j - \hat k, \qquad \vec b = x\hat i + y\hat j + z\hat k.$$

The first condition is the cross-product relation

$$\vec a \times \vec b = \vec c.$$

Using the determinant formula for a cross product,

$$\vec a \times \vec b \;=\; \begin{vmatrix} \hat i & \hat j & \hat k\\ 1 & 1 & 1\\ x & y & z \end{vmatrix} \;=\; \hat i(1\cdot z-1\cdot y)\;-\;\hat j(1\cdot z-1\cdot x)\;+\;\hat k(1\cdot y-1\cdot x).$$

Simplifying each component we get

$$\vec a \times \vec b = (\,z-y\,)\hat i - (\,z-x\,)\hat j + (\,y-x\,)\hat k.$$

Because this must equal $$\vec c = 0\hat i + 1\hat j -1\hat k,$$ we equate the corresponding components:

$$z-y = 0,\qquad -(z-x)=1,\qquad y-x=-1.$$

From the first equation we have $$z=y.$$ Substituting $$z=y$$ into the other two equations gives

$$x-z = 1 \quad\text{and}\quad y-x = -1.$$

Since $$z=y$$, the first becomes $$x-y=1$$; the second is already $$y-x=-1$$, so both are identical, confirming consistency. Hence

$$x-y = 1 \;\Longrightarrow\; x = y+1.$$

Letting $$y=t$$ (a parameter), we now have

$$x = t+1, \qquad y = t, \qquad z = t,$$

so that

$$\vec b = (t+1)\hat i + t\hat j + t\hat k.$$

The second given condition is the dot-product relation

$$\vec a \cdot \vec b = 1.$$

Using the formula $$\vec a \cdot \vec b = a_x b_x + a_y b_y + a_z b_z$$, we obtain

$$1(t+1) + 1(t) + 1(t) \;=\; 1.$$

That is

$$(t+1) + t + t \;=\; 1 \;\Longrightarrow\; 3t + 1 = 1 \;\Longrightarrow\; 3t = 0 \;\Longrightarrow\; t = 0.$$

Putting $$t=0$$ back into $$\vec b$$ gives

$$\vec b = 1\hat i + 0\hat j + 0\hat k = \hat i.$$

Next, we need the vector $$\vec a \times \vec c$$. Again applying the cross-product determinant,

$$\vec a \times \vec c = \begin{vmatrix} \hat i & \hat j & \hat k\\ 1 & 1 & 1\\ 0 & 1 & -1 \end{vmatrix} = \hat i(1\cdot(-1)-1\cdot1)\;-\;\hat j(1\cdot(-1)-1\cdot0)\;+\;\hat k(1\cdot1-1\cdot0).$$

This simplifies to

$$\vec a \times \vec c = (-2)\hat i + 1\hat j + 1\hat k.$$

We now wish to find the projection of $$\vec b$$ onto $$\vec a \times \vec c$$. The projection vector formula is

$$\text{proj}_{\vec v}\vec u \;=\; \frac{\vec u\cdot\vec v}{\vec v\cdot\vec v}\,\vec v.$$

Here $$\vec u=\vec b$$ and $$\vec v=\vec a \times \vec c$$. First, compute the needed dot products.

$$\vec b \cdot (\vec a \times \vec c) = (1,0,0)\cdot(-2,1,1) = 1\cdot(-2)+0\cdot1+0\cdot1 = -2.$$

Next, compute the squared magnitude of $$\vec a \times \vec c$$:

$$|\vec a \times \vec c|^2 = (-2)^2 + 1^2 + 1^2 = 4 + 1 + 1 = 6.$$

Therefore the projection vector is

$$\vec p = \frac{-2}{6}\,(\,-2,\,1,\,1) = -\frac13(\,-2,\,1,\,1) = \left(\frac23,\,-\frac13,\,-\frac13\right).$$

Its length $$l$$ is obtained from

$$l^2 = \left(\frac23\right)^2 + \left(-\frac13\right)^2 + \left(-\frac13\right)^2 = \frac49 + \frac19 + \frac19 = \frac69 = \frac23.$$

Hence

$$l^2 = \frac23 \quad\Longrightarrow\quad 3l^2 = 3\left(\frac23\right) = 2.$$

So, the answer is $$2$$.