To find the distance $d$ from the point $P(3, 1, -4)$ to the plane defined by the equation $4x - y - 3z = 1$, we can use the formula:

$d = \frac{|Ax_0 + By_0 + Cz_0 + D|}{\sqrt{A^2 + B^2 + C^2}}$

Here, $A = 4$, $B = -1$, $C = -3$, and $D = -1$ (from rearranging the plane equation to the form $Ax + By + Cz + D = 0$). The coordinates of the point are $(x_0, y_0, z_0) = (3, 1, -4)$.

Now, substitute the values into the formula:

1. Calculate the numerator: $|4(3) - 1(1) - 3(-4) - 1| = |12 - 1 + 12 - 1| = |22| = 22$

2. Calculate the denominator: $\sqrt{4^2 + (-1)^2 + (-3)^2} = \sqrt{16 + 1 + 9} = \sqrt{26}$

Now, plug these into the distance formula:

$d = \frac{22}{\sqrt{26}} \approx \frac{22 \sqrt{26}}{26} = \frac{11 \sqrt{26}}{13}$

So, the distance from the point to the plane is:

$d = \frac{22}{\sqrt{26}} \approx 4.3$

Would you like further details or have any questions? Here are some related questions:

1. How do you derive the equation of a plane from three points?
2. Can you explain how the distance formula changes in different dimensions?
3. What are some applications of finding the distance from a point to a plane?
4. How would the distance change if the plane's coefficients were altered?
5. Can you find the distance from a point to a line in 3D?

Tip: Always double-check your calculations by substituting back into the original equation.