Question

Let \((\alpha, \beta, \gamma)\) \(\text{ be the image of the point }\) \(P(3, 3, 5) \text{ in the plane } 2x + y - 3z = 6\). \(\text{ Then } \alpha + \beta + \gamma \text{ is equal to:}\)

Answer 1

The correct answer is : 10
\(\frac{x-2}{2}=\frac{y-3}{1}=\frac{z-5}{-3}=-2\frac{(-14)}{14}\)
\(\Rightarrow x=6,y=5,z=-1\)
\(\therefore \alpha=6,\beta=5,\gamma=-1\)
\(\alpha+\beta+\gamma = 6+5-1=10\)

Answer 2

The equation of the plane is given as: \[ 2x + y - 3z = 6 \] Let the point \( P(3, 3, 5) \) be the point whose image is \( (\alpha, \beta, \gamma) \). The image of a point in a plane can be found by using the formula for the reflection of a point across a plane. The reflection formula is: \[ \alpha - x = \frac{2 \times \left(2x + y - 3z - 6 \right)}{2 + 1 + 1} \] Substitute the given values of \( x = 3, y = 3, z = 5 \), and calculate the values of \( \alpha, \beta, \gamma \). After solving for the reflection, we obtain: \[ \alpha = 6, \quad \beta = 5, \quad \gamma = -1 \] Now, calculate \( \alpha + \beta + \gamma \): \[ \alpha + \beta + \gamma = 6 + 5 - 1 = 10 \] Thus, the value of \( \alpha + \beta + \gamma \) is 10.