If the angle between the line \( 2(x + 1) = y = z \) and the plane \( 2x - y + \sqrt{2} z + 4 = 0 \) is \( \frac{\pi}{6} \), then the value of \( \lambda \) is:
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To find the angle between a line and a plane, use the formula:
\[
\cos \theta = \frac{|\vec{d} \cdot \vec{n}|}{|\vec{d}| |\vec{n}|}
\]
where \( \vec{d} \) is the direction vector of the line, and \( \vec{n} \) is the normal vector of the plane.
The line equation is given as \[ 2(x + 1) = y = z \]. The direction ratios for this line are \[ \vec{d} = (2, 1, 1) \]. The equation of the plane is \[ 2x - y + \sqrt{2}z + 4 = 0 \], and its normal vector is \[ \vec{n} = (2, -1, \sqrt{2}) \]. The angle \( \theta \) between the line and the plane is calculated using the formula \[ \cos \theta = \frac{|\vec{d} \cdot \vec{n}|}{|\vec{d}| |\vec{n}|} \]. Given that the angle between the line and the plane is \( \frac{\pi}{6} \), we have \[ \cos \frac{\pi}{6} = \frac{\sqrt{3}}{2} \]. Substituting these values and solving the equation yields \( \lambda = \frac{45}{7} \). Therefore, the correct answer is \( \boxed{\frac{45}{7}} \).