Step 1: Understanding the Concept:
The distance of a point from a coordinate axis is found by projecting the point onto the plane perpendicular to that axis. For instance, the distance from $P(x,y,z)$ to the x-axis is the distance from $P$ to its projection $(x, 0, 0)$. The sum of the squares of these distances relates directly to the squared distance from the origin.
Step 2: Key Formula or Approach:
1. Distance from $P(x,y,z)$ to the x-axis: $d_x = \sqrt{y^2 + z^2}$.
2. Distance from $P(x,y,z)$ to the y-axis: $d_y = \sqrt{x^2 + z^2}$.
3. Distance from $P(x,y,z)$ to the z-axis: $d_z = \sqrt{x^2 + y^2}$.
4. Distance from $P(x,y,z)$ to the origin: $d_O = \sqrt{x^2 + y^2 + z^2}$.
Step 3: Detailed Explanation:
Calculate the squares of the distances from the coordinate axes:
Square of distance from x-axis: $d_x^2 = y^2 + z^2$
Square of distance from y-axis: $d_y^2 = x^2 + z^2$
Square of distance from z-axis: $d_z^2 = x^2 + y^2$
The problem states that the sum of these squared distances is 242:
\[ (y^2 + z^2) + (x^2 + z^2) + (x^2 + y^2) = 242 \]
Combine like terms:
\[ 2x^2 + 2y^2 + 2z^2 = 242 \]
Factor out the 2:
\[ 2(x^2 + y^2 + z^2) = 242 \]
Divide by 2:
\[ x^2 + y^2 + z^2 = 121 \]
The expression $x^2 + y^2 + z^2$ represents the square of the distance of point $P$ from the origin $(0,0,0)$.
Let the distance from the origin be $D$. Then $D^2 = x^2 + y^2 + z^2$.
\[ D^2 = 121 \]
Taking the square root (distance must be positive):
\[ D = \sqrt{121} = 11 \]
Step 4: Final Answer:
The distance of the point P from the origin is 11 units.