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 $\det\begin{pmatrix}1 & 0 & 1 \\ \alpha & \beta & \gamma \\ a & b & c\end{pmatrix}$ is equal to:

Step 1: Represent points on lines L1 and L2

First, we represent the general points on lines $L_1$ and $L_2$. For $L_1$, we set $\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}=\lambda_1$. This gives us the coordinates of point $A$ in terms of $\lambda_1$. For $L_2$, we have $x-6 = y$ and $y = z-4$. Let $y = \lambda_2$. Then $x = \lambda_2+6$ and $z = \lambda_2+4$. So, the coordinates of point $B$ are $(\lambda_2+6, \lambda_2, \lambda_2+4)$.

Step 2: Use collinearity of P, A, B to find parameters

Since points $P(4{,}1{,}0)$, $A(\alpha,\beta,\gamma)$, and $B(a,b,c)$ are collinear, the direction ratios of $PA$ and $PB$ must be proportional. We can write this as $\frac{\alpha-4}{a-4} = \frac{\beta-1}{b-1} = \frac{\gamma-0}{c-0}$. Substituting the expressions for $A$ and $B$ in terms of $\lambda_1$ and $\lambda_2$ into this proportionality, we get a system of equations.

Step 3: Solve for parameters $\lambda_1$ and $\lambda_2$

From the proportionality of direction ratios, we get two equations: $$(2\lambda_1-3)(3\lambda_1+2-\lambda_2) = (3\lambda_1+1)(2\lambda_1-\lambda_2-5)$$ and $$(3\lambda_1+1)(4\lambda_1-\lambda_2-1) = (4\lambda_1+3)(3\lambda_1+2-\lambda_2)$$ . Solving these simultaneous equations yields $\lambda_1 = -1$ and $\lambda_2 = 3$.

Step 4: Find coordinates of A and B

Now that we have the values of $\lambda_1$ and $\lambda_2$, we can substitute them back into the expressions for points $A$ and $B$. For $A$, using $\lambda_1 = -1$, we get $A = (2(-1)+1, 3(-1)+2, 4(-1)+3) = (-1, -1, -1)$. For $B$, using $\lambda_2 = 3$, we get $B = (3+6, 3, 3+4) = (9, 3, 7)$.

Step 5: Calculate the determinant

Finally, we need to calculate the determinant of the given matrix using the coordinates of $A(\alpha,\beta,\gamma) = (-1,-1,-1)$ and $B(a,b,c) = (9{,}3{,}7)$. The determinant is $1((-1)(7) - (-1)(3)) - 0((-1)(7) - (-1)(9)) + 1((-1)(3) - (-1)(9))$. This simplifies to $1(-7+3) + 1(-3+9) = 1(-4) + 1(6) = -4+6 = 2$.

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