Find the distance from the point to the given plane.(1, −4, 6),    3x + 2y + 6z = 5

To find the distance from a point to a plane, we can use the formula:

distance = |Ax + By + Cz + D| / sqrt(A^2 + B^2 + C^2)

where (x, y, z) is the coordinates of the point, and the equation of the plane is Ax + By + Cz + D = 0.

In this case, the equation of the plane is 3x + 2y + 6z = 5, which can be rewritten as 3x + 2y + 6z - 5 = 0.

The coordinates of the point are (1, -4, 6).

Substituting these values into the formula, we have:

distance = |3(1) + 2(-4) + 6(6) - 5| / sqrt(3^2 + 2^2 + 6^2)

Simplifying the equation, we get:

distance = |3 - 8 + 36 - 5| / sqrt(9 + 4 + 36)

distance = |26| / sqrt(49)

distance = 26 / 7

Therefore, the distance from the point (1, -4, 6) to the plane 3x + 2y + 6z = 5 is 26/7.