Let $\vec{a}, \vec{b}, \vec{c}$ be three mutually perpendicular vectors of equal magnitude. We define a new vector $\vec{r} = \vec{a} + \vec{b} + \vec{c}$. To show that $\vec{r}$ is equally inclined to $\vec{a}, \vec{b}, \vec{c}$, we must prove that the angle between $\vec{r}$ and each vector is the same.

Since the vectors are mutually perpendicular, their dot products are zero:

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

Also, let the equal magnitudes be $|\vec{a}| = |\vec{b}| = |\vec{c}| = x$. We first find the dot product of $\vec{r}$ with $\vec{a}$:

$$\vec{r} \cdot \vec{a} = (\vec{a} + \vec{b} + \vec{c}) \cdot \vec{a} = \vec{a} \cdot \vec{a} + \vec{b} \cdot \vec{a} + \vec{c} \cdot \vec{a} = |\vec{a}|^2 + 0 + 0 = x^2$$

Similarly, calculating for the other vectors, we get $\vec{r} \cdot \vec{b} = x^2$ and $\vec{r} \cdot \vec{c} = x^2$. Next, we find the magnitude of the vector $\vec{r}$. We have:

$$|\vec{r}|^2 = (\vec{a} + \vec{b} + \vec{c}) \cdot (\vec{a} + \vec{b} + \vec{c}) = |\vec{a}|^2 + |\vec{b}|^2 + |\vec{c}|^2 = x^2 + x^2 + x^2 = 3x^2$$

Thus, $|\vec{r}| = \sqrt{3}x$.

Using the formula for the angle $\theta$ between two vectors,

$$\cos \theta = \frac{\vec{u} \cdot \vec{v}}{|\vec{u}| |\vec{v}|}$$

we find the cosine of the angle between $\vec{r}$ and $\vec{a}$. Let $\alpha$ be the angle between $\vec{r}$ and $\vec{a}$. Then,

$$\cos \alpha = \frac{x^2}{(\sqrt{3}x)(x)} = \frac{1}{\sqrt{3}}$$

Similarly, the cosines of the angles between $\vec{r}$ and $\vec{b}$, and $\vec{r}$ and $\vec{c}$ are also $\frac{1}{\sqrt{3}}$. Since the cosine values are equal, the angles are equal.

∴ The vector $\vec{a} + \vec{b} + \vec{c}$ is equally inclined to $\vec{a}, \vec{b}$ and $\vec{c}$.