Rewrite the equation of the line in vector form: (x y z) = (1 -2 3) + λ (2 1 2) , λ ∈ R

And in parametric form :

x = 1 + 2λ
y = -2 + λ = λ - 2
z = 3 + 2λ

The line is parallel to (2 1 2) and (1 -2 3) is a fixed point on the line.

Now, we are given that P(x, y, z) lies on the line.

We can now define x² + y² - z² in terms of a single variable λ.

x² + y² - z²

= (1 + 2λ)² + (λ - 2)² - (3 + 2λ)²
= 1 + 4λ + 4λ² + λ² - 4λ + 4 - 9 - 12λ - 4λ²
= λ² - 12λ - 4

To find the minimum of this, let a function f(λ) be equal to x² + y² - z² Differentiate with respect to λ and set to 0.

df/dλ = 2λ - 12

When df/dλ = 0,

2λ -12 = 0
2λ = 12
λ = 6

d²f/dλ² = 2 > 0 (minimum)


So minimum value of f(λ)

= 6² - 12(6) - 4
= 36 - 72 - 4
= -40

Answer : C