Question:

Step 1: Identify the given lines and conditions. We are given three lines: Line 1 ($L_1$): $x + 2y - 3 = 0$ Line 2 ($L_2$): $3x - y - 1 = 0$ Line 3 ($L_3$): $2kx + y - 2 = 0$ We need to find values of $|k|$ under three different conditions: Case I: Line $L_1$ is parallel to Line $L_3$. Case II: Line $L_2$ is parallel to Line $L_3$. Case III: Lines $L_1$, $L_2$, and $L_3$ are concurrent. Finally, we need to sum these absolute values of $k$. Step 2: Calculate $|k|$ for Case I (Lines $L_1$ and $L_3$ are parallel). For two lines $A_1x + B_1y + C_1 = 0$ and $A_2x + B_2y + C_2 = 0$ to be parallel, the condition is $\frac{A_1}{A_2} = \frac{B_1}{B_2}$. For $L_1: x + 2y - 3 = 0$, we have $A_1=1$, $B_1=2$. For $L_3: 2kx + y - 2 = 0$, we have $A_3=2k$, $B_3=1$. Applying the parallel condition: $$ \frac{1}{2k} = \frac{2}{1} $$ $$ 1 = 4k $$ $$ k = \frac{1}{4} $$ Therefore, for Case I, $|k| = \left|\frac{1}{4}\right| = 0.25$. Step 3: Calculate $|k|$ for Case II (Lines $L_2$ and $L_3$ are parallel). Using the same parallel condition for $L_2$ and $L_3$: For $L_2: 3x - y - 1 = 0$, we have $A_2=3$, $B_2=-1$. For $L_3: 2kx + y - 2 = 0$, we have $A_3=2k$, $B_3=1$. Applying the parallel condition: $$ \frac{3}{2k} = \frac{-1}{1} $$ $$ 3 = -2k $$ $$ k = -\frac{3}{2} $$ Therefore, for Case II, $|k| = \left|-\frac{3}{2}\right| = 1.5$. Step 4: Calculate $|k|$ for Case III (Lines $L_1$, $L_2$, and $L_3$ are concurrent). For three lines $A_1x + B_1y + C_1 = 0$, $A_2x + B_2y + C_2 = 0$, and $A_3x + B_3y + C_3 = 0$ to be concurrent, the determinant of their coefficients must be zero: $$ \begin{vmatrix} A_1 & B_1 & C_1 \\ A_2 & B_2 & C_2 \\ A_3 & B_3 & C_3 \end{vmatrix} = 0 $$ Using the coefficients of $L_3$, $L_1$, and $L_2$ in that order as shown in the original solution's determinant: $L_3: 2kx + y - 2 = 0 \implies (2k, 1, -2)$ $L_1: x + 2y - 3 = 0 \implies (1, 2, -3)$ $L_2: 3x - y - 1 = 0 \implies (3, -1, -1)$ Setting the determinant to zero: $$ \begin{vmatrix} 2k & 1 & -2 \\ 1 & 2 & -3 \\ 3 & -1 & -1 \end{vmatrix} = 0 $$ Expand the determinant: $$ 2k((2)(-1) - (-3)(-1)) - 1((1)(-1) - (-3)(3)) + (-2)((1)(-1) - (2)(3)) = 0 $$ $$ 2k(-2 - 3) - 1(-1 + 9) - 2(-1 - 6) = 0 $$ $$ 2k(-5) - 1(8) - 2(-7) = 0 $$ $$ -10k - 8 + 14 = 0 $$ $$ -10k + 6 = 0 $$ $$ 10k = 6 $$ $$ k = \frac{6}{10} = \frac{3}{5} $$ Therefore, for Case III, $|k| = \left|\frac{3}{5}\right| = 0.6$. Step 5: Calculate the sum of the absolute values of $k$. Sum the values of $|k|$ found in the three cases: Sum $= |k|_{\text{Case I}} + |k|_{\text{Case II}} + |k|_{\text{Case III}}$ Sum $= 0.25 + 1.5 + 0.6$ Sum $= 2.35$ The final answer is $\boxed{2.35}$.