Let $$\vec{a} = 4\hat{i} + 3\hat{j}$$ and $$\vec{b} = 3\hat{i} - 4\hat{j} + 5\hat{k}$$ and $$\vec{c}$$ is a vector such that $$\vec{c} \cdot (\vec{a} \times \vec{b}) + 25 = 0$$, $$\vec{c} \cdot (\hat{i} + \hat{j} + \hat{k}) = 4$$ and projection of $$\vec{c}$$ on $$\vec{a}$$ is $$1$$, then the projection of $$\vec{c}$$ on $$\vec{b}$$ equals:

Given vectors $$\vec{a} = 4\hat{i} + 3\hat{j}$$ and $$\vec{b} = 3\hat{i} - 4\hat{j} + 5\hat{k}$$. Let $$\vec{c} = x\hat{i} + y\hat{j} + z\hat{k}$$.

The cross product $$\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 4 & 3 & 0 \\ 3 & -4 & 5 \end{vmatrix} = 15\hat{i} - 20\hat{j} - 25\hat{k}$$.

Since $$\vec{c} \cdot (\vec{a} \times \vec{b}) + 25 = 0$$, we have $$15x - 20y - 25z + 25 = 0$$ which simplifies to $$3x - 4y - 5z = -5$$.

The condition $$\vec{c} \cdot (\hat{i} + \hat{j} + \hat{k}) = 4$$ yields $$x + y + z = 4$$.

The projection of $$\vec{c}$$ onto $$\vec{a}$$ being 1 gives $$\frac{\vec{c} \cdot \vec{a}}{|\vec{a}|} = 1$$, so $$\frac{4x + 3y}{5} = 1$$ and hence $$4x + 3y = 5$$.

To solve for $$x,y,z$$, first write $$z = 4 - x - y$$ and substitute into $$3x - 4y - 5z = -5$$, giving $$3x - 4y - 5(4 - x - y) = -5$$, which simplifies to $$8x + y = 15$$.

From $$4x + 3y = 5$$ we get $$y = \frac{5 - 4x}{3}$$. Substituting into $$8x + y = 15$$ gives $$8x + \frac{5 - 4x}{3} = 15$$, leading to $$24x + 5 - 4x = 45$$, so $$20x = 40$$ and $$x = 2$$. Then $$y = \frac{5 - 8}{3} = -1$$ and $$z = 4 - 2 + 1 = 3$$.

Thus $$\vec{c} = 2\hat{i} - \hat{j} + 3\hat{k}$$.

The projection of $$\vec{c}$$ on $$\vec{b}$$ is $$\frac{\vec{c} \cdot \vec{b}}{|\vec{b}|} = \frac{6 + 4 + 15}{\sqrt{9+16+25}} = \frac{25}{\sqrt{50}} = \frac{25}{5\sqrt{2}} = \frac{5}{\sqrt{2}}$$, so the answer is $$\boxed{\frac{5}{\sqrt{2}}}$$, which is Option A.

The recorded answer is code 11. Our answer matches Option A. Saving for review.