Given:
The equation of the plane is given by:
\[ \left|\begin{array}{ccc|c} x-1 & y-2 & z+5 & 0 \\ 1 & 2 & 2 & \\ 1 & -3 & 7 & \end{array}\right| = 0 \]
The equation of the plane is:
\[ 4x - y - z = 7 \]
Solving for:
\[ \frac{\alpha + 1}{4} = \frac{\beta - 3}{-1} = \frac{\gamma - 4}{-1} = \frac{-2(-4 - 3 - 4 - 7)}{16 + 1 + 1} = 2 \]
Values of:
\[ \alpha = 7, \quad \beta = 1, \quad \gamma = 2 \]
Finally, the sum:
\[ \alpha + \beta + \gamma = 10 \quad (\text{Option 2}) \]
For \(a, b \in \mathbb{Z}\) and \(|a - b| \leq 10\), let the angle between the plane \(P: ax + y - z = b\) and the line \(L: x - 1 = a - y = z + 1\) be \(\cos^{-1}\left(\frac{1}{3}\right)\). If the distance of the point \((6, -6, 4)\) from the plane \(P\) is \(3\sqrt{6}\), then \(a^4 + b^2\) is equal to:
Let P₁ be the plane 3x-y-7z = 11 and P₂ be the plane passing through the points (2,-1,0), (2,0,-1), and (5,1,1). If the foot of the perpendicular drawn from the point (7,4,-1) on the line of intersection of the planes P₁ and P₂ is (α, β, γ), then a + ẞ+ y is equal to