] Find the point where the line passing through 𝑃(2, 3, −2) and parallel to the vector 𝒗 = 2𝒋 − 2𝒌 intersects the plane 6𝑥 + 3𝑦 − 4𝑧 = −12. Also verify your answer by Geogebra.

The first step is to find the equation of the line passing through the point P(2, 3, -2) and parallel to the vector v = 2j - 2k.

The equation of a line in 3D space is given by:

r = a + tb

where:

• r is the position vector of a general point on the line,
• a is the position vector of a specific point on the line,
• t is a scalar parameter,
• b is a vector parallel to the line.

Given that the line passes through P(2, 3, -2), we can say that a = 2i + 3j - 2k.

Given that the line is parallel to the vector v = 2j - 2k, we can say that b = 2j - 2k.

So, the equation of the line is:

r = (2i + 3j - 2k) + t(2j - 2k)

Next, we need to find where this line intersects the plane 6x + 3y - 4z = -12.

The equation of a plane in 3D space is given by:

ax + by + cz = d

where:

• (a, b, c) is the normal vector to the plane,
• d is the distance from the origin to the plane along the direction of the normal.

So, we can rewrite the equation of the plane as:

6x + 3y - 4z + 12 = 0

To find the intersection of the line and the plane, we substitute the parametric equations of the line into the equation of the plane and solve for t.

Then, we substitute this value of t back into the parametric equations of the line to find the coordinates of the intersection point.

As for verifying the answer using Geogebra, you would need to plot the line and the plane in the 3D Graphics view and visually check where they intersect. You can also use the Intersect tool to find the intersection point.