The image shows a parametric equation of a plane:

$\mathbf{r} = (2, 1, 4) + s(1, -2, 1) + t(1, -4, 3)$

You are tasked with converting this parametric form to the scalar (Cartesian) form of the plane's equation. You've already identified the point $\mathbf{r_0} = (2, 1, 4)$ and the normal vector $\mathbf{n} = (-2, -2, -2)$.

To convert the parametric form to scalar form, the equation of the plane can be written as:

$\mathbf{n} \cdot (\mathbf{r} - \mathbf{r_0}) = 0$

Given $\mathbf{n} = (-2, -2, -2)$ and $\mathbf{r_0} = (2, 1, 4)$, we substitute into the general form of the plane equation:

$-2(x - 2) - 2(y - 1) - 2(z - 4) = 0$

Simplifying:

$-2x + 4 - 2y + 2 - 2z + 8 = 0$ $-2x - 2y - 2z + 14 = 0$ Dividing the entire equation by $-2$:

$x + y + z - 7 = 0$

Thus, the scalar form of the equation is:

$x + y + z = 7$

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

1. How is the normal vector to a plane determined from the parametric form?
2. Why does the dot product of the normal vector and a point on the plane yield zero?
3. What is the geometric meaning of the point-normal form of a plane?
4. How can parametric equations be used in vector calculus?
5. How does this method generalize to finding the equation of a plane in 3D space?

Tip: When converting parametric forms to scalar forms, always check that the vector used as the normal is perpendicular to both direction vectors in the parametric equation.