Let $\vec{a} = 4\hat{i} + 3\hat{j}$, $\vec{b} = 3\hat{i} - 4\hat{j} + 5\hat{k}$, and $\vec{c} = x\hat{i} +…

Given:

Let $\vec{a} = 4 \hat{i} + 3 \hat{j}$ and $\vec{b} = 3 \hat{i} - 4 \hat{j} + 5 \hat{k}$.

Then,

$\vec{a} \times \vec{b} = \left| \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 4 & 3 & 0 \\ 3 & -4 & 5 \end{matrix} \right|$

Which simplifies to:

$\vec{a} \times \vec{b} = 15 \hat{i} - 20 \hat{j} - 25 \hat{k}$.

Let $\vec{c} = x \hat{i} + y \hat{j} + z \hat{k}$.

Then,

$\vec{c} \cdot (\vec{a} \times \vec{b}) + 25 = 0$

Which simplifies to:

$15x - 20y - 25z + 25 = 0$

So,

$3x - 4y - 5z = -5$ (Equation 1)

Also,

$\vec{c} \cdot (\hat{i} + \hat{j} + \hat{k}) = 4$

Which simplifies to:

$x + y + z = 4$ (Equation 2)

And, the projection of $\vec{c}$ on $\vec{a}$ is:

$\frac{\vec{c} \cdot \vec{a}}{|\vec{a}|} = 1$

Which simplifies to:

$4x + 3y = 5$ (Equation 3)

Solving Equations 1, 2, and 3, we get:

$x = 2$, $y = -1$, $z = 3$

So,

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

Finally, the projection of $\vec{c}$ on $\vec{b}$ is:

$\frac{\vec{c} \cdot \vec{b}}{|\vec{b}|} = \frac{2 \hat{i} - \hat{j} + 3 \hat{k} \cdot 3 \hat{i} - 4 \hat{j} + 5 \