<p>It is given that lines <equation>L_1: \frac{x+1}{2} = \frac{y-2}{-1} = \frac{z-1}{1}</equation> and <equation>L_2: \frac{x+2}{a} = \frac{y+1}{5-a} = \frac{z+1}{1}</equation> are coplanar. Find the value of <equation>a</equation>.</p>
Step-by-Step Solution
Key Concept: Two lines are coplanar if and only if the scalar triple product of the vector connecting a point on one line to a point on the other line with the direction vectors of both lines equals zero.
Solution: For two lines to be coplanar, the condition is: <equation>\begin{vmatrix} a & 5-a & 1 \\ 2 & -1 & 1 \\ -1 & -3 & -2 \end{vmatrix} = 0</equation> Expanding the determinant: <equation>-1(5-a+1) + 3(a-2) - 2(-a-10+2a) = 0</equation> <equation>-6+a + 3a-6 + 2a+20 = 0</equation> <equation>6a + 8 = 0</equation> <equation>a = -4</equation>
Correct Answer: -4