Final answer:
The distance from the point (1, -2, 4) to the plane 3x+2y+6z = 5 is calculated using the point-to-plane distance formula, resulting in a distance of 18/7 units.
Step-by-step explanation:
To find the distance from the point (1,-2,4) to the plane 3x+2y+6z = 5, we can use the point-to-plane distance formula. This formula states that the distance D from a point (x1,y1,z1) to the plane Ax+By+Cz+D=0 is given by:
D = |Ax1+By1+Cz1+D| / √(A2+B2+C2)
In this case, A=3, B=2, C=6, and D=-5. Substituting the point (1, -2, 4), we get:
D = |3(1) + 2(-2) + 6(4) - 5| / √(32 + 22 + 62)
D = |3 - 4 + 24 - 5| / √(9 + 4 + 36)
D = |18| / √49
D = 18 / 7
Thus, the distance from the point to the plane is 18/7 units.