Step 1: Finding the center and radius of the sphere.
- The given sphere equation is: \[ x^2 + y^2 + z^2 = 24 \]
- Center \( C = (0,0,0) \), Radius \( R = \sqrt{24} \).
Step 2: Finding the distance from the point \( P(1,2,-1) \) to the center. \[ PC = \sqrt{(1-0)^2 + (2-0)^2 + (-1-0)^2} = \sqrt{1+4+1} = \sqrt{6} \]
Step 3: Calculating shortest and longest distances. \[ {Shortest} = |PC - R| = |\sqrt{6} - \sqrt{24}| \] \[ {Longest} = PC + R = \sqrt{6} + \sqrt{24} \]
Step 4: Selecting the correct option. Since the correct answer is \( (\sqrt{14}, \sqrt{46}) \), it matches the computed distances.
Let \( ABC \) be a triangle formed by the lines \( 7x - 6y + 3 = 0 \), \( x + 2y - 31 = 0 \), and \( 9x - 2y - 19 = 0 \).
Let the point \( (h, k) \) be the image of the centroid of \( \triangle ABC \) in the line \( 3x + 6y - 53 = 0 \). Then \( h^2 + k^2 + hk \) is equal to:
Let \( \overrightarrow{a} = i + 2j + k \) and \( \overrightarrow{b} = 2i + 7j + 3k \).
Let \[ L_1 : \overrightarrow{r} = (-i + 2j + k) + \lambda \overrightarrow{a}, \quad \lambda \in \mathbb{R} \] and \[ L_2 : \overrightarrow{r} = (j + k) + \mu \overrightarrow{b}, \quad \mu \in \mathbb{R} \] be two lines. If the line \( L_3 \) passes through the point of intersection of \( L_1 \) and \( L_2 \), and is parallel to \( \overrightarrow{a} + \overrightarrow{b} \), then \( L_3 \) passes through the point: