Find the equation of the plane through the points (2, 1, –1)

We know that, equation of the plane passing through two points (x1, y1, z1) and (x2, y2, z2) with its normal’s direction ratios is

a(x – x1) + b(y – y1) + c(z – z1) = 0 …. (i)

Now, if the plane is passing through two points (2, 1, -1) and (-1, 3, 4) then

a(x2 -x1) + b(y2 – y1) + c(z2 – z1) = 0

a(-1 – 2) + b(3 – 1) + c(4 + 1) = 0

-3a + 2b + 5c = 0 …. (ii)

As the required plane is perpendicular to the given plane x – 2y + 4z = 10, then

1.a – 2.b + 4.c = 10 …. (iii)

On solving (ii) and (iii) we get,

$\frac{a}{8+10}=\frac{-b}{-12-5}=\frac{c}{6-2}=\lambda$

So, a = 18λ, b = 17λ and c = 4λ

Thus, the required plane is

18λ (x – 2) + 17λ (y – 1) + 4λ (z + 1) = 0

18x – 36 + 17y – 17 + 4z + 4 = 0

⇒ 18x + 17y + 4z – 49 = 0