If $\Delta = \begin{vmatrix} 1 & a & a^2 \\ 1 & b & b^2 \\ 1 & c & c^2 \end{vmatrix} = k(a-b)(b-c)(c-a)$, then $k$ is equal to

Question

Let the foot of the perpendicular from the point $ (1, 2, 4) $ on the line $ \frac{x+2}{4} = \frac{y-1}{2} = \frac{z+1}{3} $ be $ P $. Then the distance of $ P $ from the plane $ 3x + 4y + 12z + 23 = 0 $ is:

Answer 1

To find the value of $ k $ in the given determinant equation, we first need to evaluate the determinant $ \Delta $ given by:

\[ \Delta = \begin{vmatrix} 1 & a & a^2 \\ 1 & b & b^2 \\ 1 & c & c^2 \end{vmatrix} \]

The value of the determinant can be expanded as follows:

\[ \Delta = 1 \cdot \left(b \cdot c^2 - c \cdot b^2\right) - a \cdot \left(c \cdot b^2 - b \cdot c^2\right) + a^2 \cdot \left(c \cdot b - b \cdot c\right) \]

Simplifying each term:

The first term: $ b \cdot c^2 - c \cdot b^2 = bc^2 - cb^2 = bc^2 - b^2c = bc(c-b) $
The second term: $ a(c \cdot b^2 - b \cdot c^2) = a(b^2c - c^2b) = a(bc^2 - b^2c) = a(bc)(c-b) $
The third term becomes zero because: $ a^2(c \cdot b - b \cdot c) = a^2(0) = 0 $

Substituting back, we have:

\[ \Delta = bc(c-b) - a(bc)(c-b) \]

We can factor out $ (c-b) $:

\[ \Delta = (c-b)(bc - abc) \]

Simplify the expression inside the brackets:

\[ \Delta = (c-b)(b - ab)c \]

Simplifying the remaining expression within the brackets:

\[ \Delta = (c-b)c(b-a) \]

This can be rewritten due to the cyclic nature of the equations involving determinants:

\[ \Delta = (a-b)(b-c)(c-a) \]

Comparing this with the expression given in the question $ \Delta = k(a-b)(b-c)(c-a) $, we find that:

\[ k = 1 \]

Therefore, the correct answer is:

1

Answer 2