Exercise: Find an equation in general form for the plane through 2, 1,  1 that is parallel
to the plane with equation x  2 y  3z  4 .
Since the planes are parallel, the normal vector n
⃗ = [1, −2, 3] for the given plane is also a
normal vector for the new plane through 2, 1,  1 .
So the new plane has general equation
for some real number 𝑑.
𝑥 − 2𝑦 + 3𝑧 = 𝑑
We know 2, 1,  1 is a point on the plane so 2, 1,  1 must satisfy the equation of the plane.
Substituting in 2, 1,  1 , we have
2 − 2(1) + 3(−1) = 𝑑
−3 = 𝑑
Thus, a general equation for the plane is
𝑥 − 2𝑦 + 3𝑧 = −3