Given that vectors a, b, and c are mutually perpendicular, find the dot product X * Y where X = 4a − 3b + 2c and Y = 3a + 2b − 5c.

Given that vectors $\vec{a}$, $\vec{b}$, and $\vec{c}$ are mutually perpendicular, their dot products are zero:

$$\vec{a} \cdot \vec{b} = 0$$ $$\vec{a} \cdot \vec{c} = 0$$ $$\vec{b} \cdot \vec{c} = 0$$

Also, the dot product of a vector with itself gives the square of its magnitude:

$$\vec{a} \cdot \vec{a} = |\vec{a}|^2$$ $$\vec{b} \cdot \vec{b} = |\vec{b}|^2$$ $$\vec{c} \cdot \vec{c} = |\vec{c}|^2$$

Given that vectors $\vec{X}$ and $\vec{Y}$ are perpendicular, their dot product is zero:

$$\vec{X} \cdot \vec{Y} = 0$$

Substitute $\vec{X} = 4\vec{a} - 3\vec{b} + 2\vec{c}$ and $\vec{Y} = 3\vec{a} + 2\vec{b} - 5\vec{c}$:

$$\vec{X} \cdot \vec{Y} = (4\vec{a} - 3\vec{b} + 2\vec{c}) \cdot (3\vec{a} + 2\vec{b} - 5\vec{c})$$

Expand the dot product, using the mutual perpendicularity of $\vec{a}, \vec{b}, \vec{c}$:

$$\vec{X} \cdot \vec{Y} = (4\vec{a} \cdot 3\vec{a}) + (4\vec{a} \cdot 2\vec{b}) + (4\vec{a} \cdot -5\vec{c}) + (-3\vec{b} \cdot 3\vec{a}) + (-3\vec{b} \cdot 2\vec{b}) + (-3\vec{b} \cdot -5\vec{c}) + (2\vec{c} \cdot 3\vec{a}) + (2\vec{c} \cdot 2\vec{b}) + (2\vec{c} \cdot -5\vec{c})$$ $$\vec{X} \cdot \vec{Y} = 12(\vec{a} \cdot \vec{a}) + 8(\vec{a} \cdot \vec{b}) - 20(\vec{a} \cdot \vec{c}) - 9(\vec{b} \cdot \vec{a}) - 6(\vec{b} \cdot \vec{b}) + 15(\vec{b} \cdot \vec{c}) + 6(\vec{c} \cdot \vec{a}) + 4(\vec{c} \cdot \vec{b}) - 10(\vec{c} \cdot \vec{c})$$

Since $\vec{a}, \vec{b}, \vec{c}$ are mutually perpendicular, all cross-term dot products are zero:

$$\vec{X} \cdot \vec{Y} = 12|\vec{a}|^2 - 6|\vec{b}|^2 - 10|\vec{c}|^2$$

Given $|\vec{b}| = \sqrt{17}$ and $|\vec{c}| = 3$. Since $\vec{X} \cdot \vec{Y} = 0$:

$$12|\vec{a}|^2 - 6(\sqrt{17})^2 - 10(3)^2 = 0$$ $$12|\vec{a}|^2 - 6(17) - 10(9) = 0$$ $$12|\vec{a}|^2 - 102 - 90 = 0$$ $$12|\vec{a}|^2 - 192 = 0$$