It is given that,

A plane passes through the points (2, 1, -1) and (-1, 3, 4) and perpendicular to the plane x – 2y + 4z = 10.

We need to find the equation of such plane.

We know that,

The cartesian equation of a plane passing through (x₁, y₁, z₁) having direction ratios proportional to a, b, c for its normal is

a(x – x₁) + b(y – y₁) + c(z – z₁) = 0

So,

Let the equation of the plane passing through (2, 1, -1) be

a(x – 2) + b(y – 1) + c(z – (-1)) = 0

⇒ a(x – 2) + b(y – 1) + c(z + 1) = 0 …(i)

Since, it also passes through point (-1, 3, 4), just replace x, y, z by -1, 3, 4 respectively.

⇒ a(-1 – 2) + b(3 – 1) + c(4 + 1) = 0

⇒ -3a + 2b + 5c = 0 …(ii)

Since, a, b, c are direction ratios and this plane is perpendicular to the plane x – 2y + 4z = 10, just replace x, y, z by a, b, c (neglecting 10) respectively and equate to 0. So, we get

a – 2b + 4c = 0 …(iii)

If we need to solve two equations x₁a + y₁b + z₁c = 0 and x₂a + y₂b + z₂c = 0, the formula is:

Similarly, solve for equations (ii) and (iii).

That is,

⇒ a = 18λ

⇒ b = 17λ

⇒ c = 4λ

Substitute these values of a, b, c in equation (i),

a(x – 2) + b(y – 1) + c(z + 1) = 0

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

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

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

⇒ 18x – 36 + 17y – 17 + 4z + 4 = 0

⇒ 18x + 17y + 4z – 36 – 17 + 4 = 0

⇒ 18x + 17y + 4z – 49 = 0

⇒ 18x + 17y + 4z = 49

Thus, equation of the required plane is 18x + 17y + 4z = 49.