Question

Let \(\alpha x+\beta y+y z=1\) be the equation of a plane passing through the point\((3,-2,5)\)and perpendicular to the line joining the points \((1,2,3)\) and \((-2,3,5)\) Then the value of \(\alpha \beta y\)is equal to ____

Hint:

When dealing with planes perpendicular to a line, always compute the direction vector of the line and use it as the normal vector of the plane.

Answer 1

Correct Answer: 6

The given equation is not the equation of a plane, as \(yz\) is present. If we consider \(y\) as \(\gamma\), the equation of the plane becomes:
\[ax + \beta y + \gamma z = 1.\]
Step 1: Find the Normal Vector of the Plane
The plane is perpendicular to the line joining the points \((1, 2, 3)\) and \((-2, 3, 5)\). The direction vector of the line is:
\[\vec{d} = (-2 - 1)\mathbf{i} + (3 - 2)\mathbf{j} + (5 - 3)\mathbf{k} = -3\mathbf{i} + \mathbf{j} + 2\mathbf{k}.\]
The normal vector of the plane is parallel to \(\vec{d}\), so:
\[\vec{n} = 3\mathbf{i} - \mathbf{j} - 2\mathbf{k}.\]
Step 2: Equation of the Plane
The equation of the plane passing through \((3, -2, 5)\) is:
\[3x - y - 2z + \lambda = 0.\]
Substituting \((3, -2, 5)\) into the plane equation:
\[3(3) - (-2) - 2(5) + \lambda = 0,\]
\[9 + 2 - 10 + \lambda = 0,\]
\[\lambda = -1.\]
Thus, the equation of the plane is:
\[3x - y - 2z = 1.\]
Step 3: Calculate \(\alpha \beta \gamma\)
Comparing the plane equation \(3x - y - 2z = 1\) with \(ax + \beta y + \gamma z = 1\), we have:
\[a = 3, \quad \beta = -1, \quad \gamma = -2.\]
Therefore:
\[\alpha \beta \gamma = 3(-1)(-2) = 6.\]
Conclusion
\[\alpha \beta \gamma = 6.\]