Animated Solution for Mathematics - Three Dimensional Geometry: The equation of the plane containing the line $2x-5y+z=3$; $x+y+4z=5$, and parallel to the plane, $x+3y+6z=1$, is:

When two planes, $P_1:2x-5y+z-3=0$ and $P_2:x+y+4z-5=0$, intersect, they create a line $L$. We could spend time finding the parametric equation of this line, but that is a trap for the unwary.

Instead, we use the Family of Planes theorem. Think of this line $L$ as a hinge on a door. By varying a parameter $\lambda$, we can rotate a plane around this hinge to any angle we desire.

The equation for this entire family is simply $P_1+\lambda P_2=0$. It is a beautiful, compact way to represent an infinite number of planes without ever needing to know the coordinates of the line itself. We write this as:

Now, we apply the constraint. We are told our new plane must be parallel to $x+3y+6z=1$. In the language of vectors, parallelism is synonymous with proportionality.

If two planes are parallel, their normal vectors must point in the same direction. The normal vector of our family plane is derived from the coefficients of $x$, $y$, and $z$. After grouping the terms, we get:

The normal vector is $\vec{n}=((2+\lambda ),(-5+\lambda ),(1+4\lambda ))$. For this to be parallel to the target plane with normal ${\vec{n}}_{target}=(1,3,6)$, the ratios of their components must be equal:

Now, we solve for $\lambda$. Taking the first two ratios, we have:

Cross-multiplying gives us $3(2+\lambda )=-5+\lambda$, which simplifies to $6+3\lambda =-5+\lambda$. Solving this linear equation, we find $2\lambda =-11$, or $\lambda =-\frac{11}{2}$.

We substitute $\lambda =-\frac{11}{2}$ back into our grouped equation:

Simplifying these terms yields:

To make this look elegant, we multiply the entire equation by $-\frac{2}{7}$, resulting in $x+3y+6z-7=0$.

Rearranging, we arrive at our final answer: