Let a line passing through the point $$(4, 1, 0)$$ intersect the line $$L_1 : \frac{x-1}{2} = \frac{y-2}{3} = \frac{z-3}{4}$$ at the point $$A(\alpha, \beta, \gamma)$$ and the line $$L_2 : x - 6 = y = -z + 4$$ at the point $$B(a, b, c)$$. Then $$\begin{vmatrix} 1 & 0 & 1 \\ \alpha & \beta & \gamma \\ a & b & c \end{vmatrix}$$ is equal to

The given lines are
  $$L_1:\;\dfrac{x-1}{2}= \dfrac{y-2}{3}= \dfrac{z-3}{4}$$
  $$L_2:\;x-6 = y = -z+4$$

Write them in parametric form.

For $$L_1$$ let the parameter be $$t$$:
  $$x = 1+2t,\;y = 2+3t,\;z = 3+4t\;.\tag{-1}$$

Hence the intersection point with $$L_1$$ is
  $$A(\alpha,\beta,\gamma)=\bigl(1+2t,\;2+3t,\;3+4t\bigr).$$

For $$L_2$$ write $$y=\lambda$$. From $$x-6=y$$ we get $$x = 6+\lambda$$, and from $$y = -z+4$$ we get $$z = 4-\lambda$$. So

$$B(a,b,c)=\bigl(6+\lambda,\;\lambda,\;4-\lambda\bigr).\tag{-2}$$

The required line passes through the given point $$P(4,1,0)$$ and the two points $$A$$ and $$B$$, so the three points are collinear. Therefore the vectors $$\overrightarrow{PA}$$ and $$\overrightarrow{PB}$$ must be parallel.

Compute the two vectors:

$$\overrightarrow{PA} = (\,(1+2t)-4,\;(2+3t)-1,\;(3+4t)-0) = (-3+2t,\;1+3t,\;3+4t).$$

$$\overrightarrow{PB} = (\,(6+\lambda)-4,\;\lambda-1,\;(4-\lambda)-0) = (2+\lambda,\;\lambda-1,\;4-\lambda).$$

Parallelism means the three component ratios are equal to a common constant $$k$$:

$$\dfrac{-3+2t}{2+\lambda}= \dfrac{1+3t}{\,\lambda-1\,}= \dfrac{3+4t}{\,4-\lambda\,}=k.\tag{-3}$$

First equate the ratios formed by the first two components:

$$( -3+2t)(\lambda-1) = (1+3t)(2+\lambda).\tag{-4}$$

Next equate the ratios formed by the first and the third components:

$$( -3+2t)(4-\lambda) = (3+4t)(2+\lambda).\tag{-5}$$

Expand $$( -4)$$:

$$( -3+2t)\lambda -(-3+2t) = (1+3t)2 + (1+3t)\lambda$$
$$(-3+2t)\lambda +3 -2t = 2 + 6t + (1+3t)\lambda.$$

Collecting terms gives $$(-4-t)\lambda +1 -8t = 0.$$ Hence

$$\lambda(-4-t)=8t-1.\tag{-6}$$

Expand $$( -5)$$:

$$( -3+2t)(4-\lambda) = (3+4t)(2+\lambda)$$
$$-12 + 8t +3\lambda -2t\lambda = 6 + 8t + 3\lambda + 4t\lambda.$$

Simplifying yields $$-18 - 6t\lambda = 0 \quad\Longrightarrow\quad t\lambda = -3.\tag{-7}$$

Use $$( -7)$$ in $$( -6)$$. From $$( -7)$$, $$\lambda= -\dfrac{3}{t} \;(t\neq 0).$$ Substitute in $$( -6)$$:

$$\left(-\dfrac{3}{t}\right)(-4-t) = 8t-1$$
$$\dfrac{12}{t} + 3 = 8t - 1.$$

Multiply by $$t$$:

$$12 + 3t = 8t^{2} - t.$$

Re-arrange:

$$8t^{2} -4t -12 = 0 \quad\Longrightarrow\quad 2t^{2}-t-3=0.$$

Quadratic formula gives $$t = \dfrac{1\pm 5}{4}\;\Longrightarrow\; t=\dfrac{3}{2}\;\text{or}\;t=-1.$$

Check each value in $$( -3)$$ to ensure a single proportionality constant $$k$$ exists.

Case 1: $$t=\dfrac{3}{2} \Longrightarrow \lambda=-2.$$br/> Then $$\overrightarrow{PA}=(0,\,5.5,\,9)$$ and $$\overrightarrow{PB}=(0,\,-3,\,6).$$ The second and third component ratios are different ($$-5.5/3\neq 9/6$$). Hence the three points are \emph{not} collinear - this root is extraneous. Case 2: $$t=-1 \Longrightarrow \lambda = 3.$$br/> Now

$$A(-1,-1,-1),\qquad B(9,3,1).$$

Vectors $$\overrightarrow{PA}=(-5,\; -2,\; -1),\quad \overrightarrow{PB}=(5,\; 2,\; 1)$$ are negatives of each other, so the points $$P,A,B$$ are indeed collinear. Thus $$t=-1,\;\lambda=3$$ is the only acceptable solution.

Fill the matrix with the obtained coordinates:

$$\begin{vmatrix} 1 & 0 & 1\\ \alpha & \beta & \gamma\\ a & b & c \end{vmatrix} = \begin{vmatrix} 1 & 0 & 1\\ -1 & -1 & -1\\ 9 & 3 & 1 \end{vmatrix}.$$

Evaluate the determinant:

$$\begin{aligned} \Delta &= 1\bigl((-1)(1) - (-1)(3)\bigr) - 0(\cdots) + 1\bigl((-1)(3) - (-1)(9)\bigr)\\[4pt] &= 1(-1 + 3) + 1(-3 + 9)\\ &= 1(2) + 1(6)\\ &= 8. \end{aligned}$$

Hence $$\begin{vmatrix} 1 & 0 & 1 \\ \alpha & \beta & \gamma \\ a & b & c \end{vmatrix}=8.$$

The correct choice is Option A (8).