Let's solve the problem step by step. Here's how we approach each part:

(a) Sketch the triangle

Since this is a 3D triangle, a direct sketch would involve plotting points in 3D space. The three points of the triangle are:

• $P(1, 1, 3)$
• $Q(4, -1, 3)$
• $R(7, 1, -2)$

Although I cannot generate a 3D sketch here, I can describe the process:

1. Axes: Draw the three coordinate axes $x$, $y$, and $z$.
2. Plot points:
   • $P = (1, 1, 3)$: Move 1 unit along the x-axis, 1 unit along the y-axis, and 3 units along the z-axis.
   • $Q = (4, -1, 3)$: Move 4 units along the x-axis, -1 unit (down) along the y-axis, and 3 units along the z-axis.
   • $R = (7, 1, -2)$: Move 7 units along the x-axis, 1 unit along the y-axis, and -2 units (down) along the z-axis.
3. Connect the points to form the triangle.

If you'd like a graphical representation, I can assist you with that as well.

(b) Find the vector and scalar equations of the plane

1. Vector equation of the plane

To find the vector equation, we need a point on the plane and two direction vectors that lie on the plane.

• Choose $P(1, 1, 3)$ as the reference point.

• Find two direction vectors using the points $Q$ and $R$:

  $\overrightarrow{PQ} = Q - P = (4 - 1, -1 - 1, 3 - 3) = (3, -2, 0)$ $\overrightarrow{PR} = R - P = (7 - 1, 1 - 1, -2 - 3) = (6, 0, -5)$

The vector equation of a plane is given by: $\mathbf{r} = \mathbf{P} + s \overrightarrow{PQ} + t \overrightarrow{PR}$ Substituting the values of $P$, $\overrightarrow{PQ}$, and $\overrightarrow{PR}$:

$\mathbf{r} = (1, 1, 3) + s(3, -2, 0) + t(6, 0, -5)$

Thus, the vector equation is: $\mathbf{r} = (1 + 3s + 6t, 1 - 2s, 3 - 5t)$

2. Scalar equation of the plane

The scalar equation of a plane is of the form: $Ax + By + Cz = D$ To find $A$, $B$, and $C$, we need the normal vector $\mathbf{n}$ to the plane. The normal vector is obtained by taking the cross product of the direction vectors $\overrightarrow{PQ}$ and $\overrightarrow{PR}$:

$\mathbf{n} = \overrightarrow{PQ} \times \overrightarrow{PR}$ $\mathbf{n} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 3 & -2 & 0 \\ 6 & 0 & -5 \end{vmatrix}$

Let's compute this:

$\mathbf{n} = \mathbf{i}((-2)(-5) - (0)(0)) - \mathbf{j}((3)(-5) - (0)(6)) + \mathbf{k}((3)(0) - (-2)(6))$ $\mathbf{n} = \mathbf{i}(10) - \mathbf{j}(-15) + \mathbf{k}(12)$ $\mathbf{n} = (10, 15, 12)$

Thus, the normal vector to the plane is $\mathbf{n} = (10, 15, 12)$.

Now, use the point $P(1, 1, 3)$ to find $D$ in the scalar equation. The scalar equation is given by: $10(x - 1) + 15(y - 1) + 12(z - 3) = 0$

Expanding this: $10x - 10 + 15y - 15 + 12z - 36 = 0$ $10x + 15y + 12z = 61$

Thus, the scalar equation of the plane is: $10x + 15y + 12z = 61$

Summary of Results:

1. Vector equation of the plane: $\mathbf{r} = (1 + 3s + 6t, 1 - 2s, 3 - 5t)$
2. Scalar equation of the plane: $10x + 15y + 12z = 61$

Would you like further details or clarifications on any part of this? Here are some related questions:

1. How is the cross product used to find the normal vector to a plane?
2. How can you interpret the scalar equation of a plane geometrically?
3. What is the difference between a vector and scalar equation in 3D geometry?
4. How do you check if a point lies on a given plane?
5. What other forms of the equation of a plane can we use in 3D space?

Tip: To verify your results, substitute the coordinates of $P$, $Q$, and $R$ into the scalar equation to ensure that they all satisfy it.