The problem asks to find the equation of a plane that passes through the point $D(1, 0, 2)$ and contains the $yz$-plane.

1. Problem Description

The problem asks to find the equation of a plane that passes through the point $D(1, 0, 2)$ and contains the $yz$-plane.

2. Solution Steps

Since the plane contains the $yz$-plane, the equation of the plane is of the form $Ax = 0$ for some constant $A$.

Since the plane passes through the point $D(1, 0, 2)$, we can substitute the coordinates of point $D$ into the equation to find the value of $A$.

$A(1) = 0$

$A = 0$

However, if $A=0$, we get $0=0$ which does not give any information.

Alternatively, since the plane contains the $yz$-plane, it must have a normal vector that is perpendicular to the $yz$-plane.

A normal vector to the $yz$-plane is $\vec{i} = (1, 0, 0)$.

The equation of the plane passing through $D(1, 0, 2)$ with normal vector $(A, B, C)$ is given by

$A(x - 1) + B(y - 0) + C(z - 2) = 0$

$Ax + By + Cz - A - 2C = 0$

Since the plane contains the $yz$-plane, the point $(0, y, z)$ must satisfy the equation $Ax + By + Cz - A - 2C = 0$ for all $y, z$. Thus,

$A(0) + By + Cz - A - 2C = 0$

$By + Cz - A - 2C = 0$

Since this holds for all $y$ and $z$, $B = 0$ and $C = 0$. Therefore $-A - 2C = 0$ implies $-A = 0$, so $A = 0$.

This again implies that $0=0$, which is not very helpful.

If a plane contains the yz-plane then its equation is of the form $Ax = 0$, which reduces to $x=0$.

We know that the plane must pass through $D(1, 0, 2)$, so substituting this point into the equation we get $A(1) = 0$, which means $A = 0$. Thus $x=0$.

However, this is the $yz$-plane. So something is missing.

Consider the points $(0,0,0)$ which is on the $yz$ plane, and the point $(1,0,2)$.

If the equation of the plane is $Ax+By+Cz+D=0$, then we require that $0 = B(y)+C(z)+D$ for all $y, z$ so $B = C = D = 0$.

Then the equation of the plane is $Ax=0$.

And the point $(1,0,2)$ must satisfy this so $A(1) = 0$ which implies $A=0$.

Since the plane contains the $yz$-plane, it must be of the form $Ax = 0$. The plane also passes through $D(1, 0, 2)$, so $x = 0$ is not the only possibility.

Since it includes the $yz$-plane, $x=0$.

Let the equation of the plane be $x = 0$. However, the point (1, 0, 2) is not on the $yz$-plane. This seems strange.

If the plane *intersects* the yz plane, then $x=0$. If the plane *contains* the yz plane, then the yz plane is part of the equation.

Then the equation can be $x=0$.

If the plane includes the yz plane and passes through (1,0,2), then the equation is of the form $Ax + D = 0$, but $B, C = 0$. The yz plane implies x=

0. But (1,0,2) implies x=

1.

3. Final Answer

The equation of the $yz$-plane is $x=0$. Since the plane must pass through $D(1,0,2)$, let the equation be $x=0$.

The plane $x=0$ contains the $yz$ plane. However, it does not pass through $D(1, 0, 2)$. Thus there seems to be an inconsistency.

Final Answer: x=0