To find the equation of a plane, we need a point on the plane and a normal vector to the plane. We are given a point on the plane: (1, 2, –3).

We are also told that the plane is normal to the straight line joining the points (–1, 3, 4) and (5, 2, –1). This means the direction vector of this line is parallel to the normal vector of the plane.

Finding the Normal Vector

The direction vector of the line joining two points \(A(x_1, y_1, z_1)\) and \(B(x_2, y_2, z_2)\) is given by the vector \(\vec{AB} = \langle x_2 - x_1, y_2 - y_1, z_2 - z_1 \rangle\).

Let the two points on the line be \(P(-1, 3, 4)\) and \(Q(5, 2, -1)\). The direction vector of the line joining P and Q is:

This direction vector \(\langle 6, -1, -5 \rangle\) is normal to the plane. So, the normal vector to the plane is \(\mathbf{n} = \langle 6, -1, -5 \rangle\). The components of the normal vector are \(a=6\), \(b=-1\), and \(c=-5\).

Equation of the Plane

The equation of a plane passing through a point \((x_0, y_0, z_0)\) with a normal vector \(\langle a, b, c \rangle\) is given by:

We have the point on the plane \((x_0, y_0, z_0) = (1, 2, -3)\) and the normal vector \(\langle a, b, c \rangle = \langle 6, -1, -5 \rangle\).

Substitute these values into the plane equation:

Simplifying the Equation

Now, let's simplify the equation by distributing and combining terms:

Combine the constant terms:

So, the equation of the plane is:

This is the equation of the plane that passes through the point (1, 2, –3) and is normal to the line joining (–1, 3, 4) and (5, 2, –1).

Verifying the Equation

Let's compare our derived equation \(6x - y - 5z - 19 = 0\) with the given options.

• Option 1: \(6x - y - 5z - 19 = 0\)
• Option 2: \(x - 2y - 3z - 17 = 0\)
• Option 3: \(6x - y + 5z + 19 = 0\)
• Option 4: \(3x - 2y - 5z - 9 = 0\)

Our derived equation matches Option 1.