Let $$\overrightarrow{a}=\widehat{i}-2\widehat{j}+3\widehat{k}, \overrightarrow{b}=2\widehat{i}+\widehat{j}-\widehat{k}, \overrightarrow{c}=\lambda \widehat{i}+\widehat{j}+\widehat{k}$$ and $$\overrightarrow{v}= \overrightarrow{a} \times \overrightarrow{b}$$. If $$\overrightarrow{v}\cdot\overrightarrow{c}=11$$ and the length of the projection of $$\overrightarrow{b}$$ on $$\overrightarrow{c}$$ is p, then $$9p^{2}$$ is equal to

Given: $$\vec{a} = \hat{i} - 2\hat{j} + 3\hat{k}$$, $$\vec{b} = 2\hat{i} + \hat{j} - \hat{k}$$, $$\vec{c} = \lambda\hat{i} + \hat{j} + \hat{k}$$, and $$\vec{v} = \vec{a} \times \vec{b}$$.

We need to find $$9p^2$$ where $$\vec{v} \cdot \vec{c} = 11$$ and $$p$$ is the length of projection of $$\vec{b}$$ on $$\vec{c}$$.

Computing the cross product $$\vec{v} = \vec{a} \times \vec{b}$$, we evaluate the determinant $$\vec{v} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -2 & 3 \\ 2 & 1 & -1 \end{vmatrix}$$ which expands as $$= \hat{i}[(-2)(-1) - (3)(1)] - \hat{j}[(1)(-1) - (3)(2)] + \hat{k}[(1)(1) - (-2)(2)]$$ then simplifies to $$= \hat{i}[2 - 3] - \hat{j}[-1 - 6] + \hat{k}[1 + 4]$$ and finally yields $$= -\hat{i} + 7\hat{j} + 5\hat{k}.$$

Using $$\vec{v} \cdot \vec{c} = 11$$ we write $$\vec{v} \cdot \vec{c} = (-1)(\lambda) + (7)(1) + (5)(1) = -\lambda + 7 + 5 = 12 - \lambda$$ so that $$12 - \lambda = 11$$, which implies $$\lambda = 1$$. Thus $$\vec{c} = \hat{i} + \hat{j} + \hat{k}$$.

The length of projection of $$\vec{b}$$ on $$\vec{c}$$ is given by $$p = \frac{|\vec{b} \cdot \vec{c}|}{|\vec{c}|},$$ with $$\vec{b} \cdot \vec{c} = (2)(1) + (1)(1) + (-1)(1) = 2 + 1 - 1 = 2$$ and $$|\vec{c}| = \sqrt{1 + 1 + 1} = \sqrt{3},$$ hence $$p = \frac{|2|}{\sqrt{3}} = \frac{2}{\sqrt{3}}.$$

Since $$p^2 = \frac{4}{3},$$ it follows that $$9p^2 = 9 \times \frac{4}{3} = 12.$$

The correct answer is Option (1): $$\boxed{12}$$.