Find the equation of the plane passing through the point (1, 2, -1) and perpendicular to the line with direction ratios 2, -1, 3.
Correct: A
The direction ratios of the normal to the plane are the same as the direction ratios of the line, which are 2, -1, 3. The equation of the plane passing through (x₁, y₁, z₁) and having direction ratios a, b, c for the normal is a(x - x₁) + b(y - y₁) + c(z - z₁) = 0. Here, (x₁, y₁, z₁) = (1, 2, -1) and a = 2, b = -1, c = 3. So, the equation is 2(x - 1) - 1(y - 2) + 3(z - (-1)) = 0, which simplifies to 2x - 2 - y + 2 + 3z + 3 = 0, or 2x - y + 3z + 3 = 0. Therefore, 2x - y + 3z = -3.