\(\frac{25}{4\sqrt3}\)
\(\frac{25\sqrt3}{2}\)
\(\frac{25}{\sqrt3}\)
\(\frac{25}{2\sqrt3}\)
The correct answer is (D) : \(\frac{25}{2\sqrt3}\)
Altitude of equilateral triangle,
\(\frac{\sqrt3l}{2}=\frac{5}{\sqrt2}\)
\(l=\frac{5\sqrt2}{\sqrt3}\)
Area of triangle
\(=\frac{\sqrt3}{4}l^2=\frac{\sqrt3}{4}.\frac{50}{3}=\frac{25}{2\sqrt3}\)
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} \text{ is equal to} $