Let P be the plane, passing through the point 1,−1,−5  and perpendicular to the line joining the points 4,1,−3 and 2,4,3.  Then the distance of P from the point 3,−2,2 is

To find the distance of the plane P from the point (3, -2, 2), we need to follow these steps:

Step 1: Find the direction vector of the line joining the points (4, 1, -3) and (2, 4, 3).

• Subtract the coordinates of the second point from the coordinates of the first point to get the direction vector: (4 - 2, 1 - 4, -3 - 3) = (2, -3, -6).

Step 2: Find the normal vector of the plane P.

• Since the plane is perpendicular to the line, the direction vector of the line is also the normal vector of the plane: (2, -3, -6).

Step 3: Use the equation of a plane to find the equation of plane P.

• The equation of a plane is given by Ax + By + Cz + D = 0, where (A, B, C) is the normal vector of the plane and (x, y, z) are the coordinates of a point on the plane.
• Substitute the coordinates of the given point (1, -1, -5) and the normal vector (2, -3, -6) into the equation to find D: 2(1) + (-3)(-1) + (-6)(-5) + D = 0 2 + 3 + 30 + D = 0 D = -35
• The equation of plane P is 2x - 3y - 6z - 35 = 0.

Step 4: Find the distance between the point (3, -2, 2) and plane P.

• Use the formula for the distance between a point and a plane: Distance = |Ax + By + Cz + D| / sqrt(A^2 + B^2 + C^2)
• Substitute the coordinates of the point (3, -2, 2) and the coefficients of the plane equation into the formula: Distance = |2(3) - 3(-2) - 6(2) - 35| / sqrt(2^2 + (-3)^2 + (-6)^2) Distance = |6 + 6 - 12 - 35| / sqrt(4 + 9 + 36) Distance = |-35| / sqrt(49) Distance = 35 / 7 Distance = 5

Therefore, the distance of plane P from the point (3, -2, 2) is 5 units.