<p>The three planes <br>
\(x - 2y + z - 1 = 0\) ...(1)<br>
\(x + 2y - 2z - 5 = 0\) ...(2)<br>
\(x + y - 2z - 7 = 0\) ...(3)<br>
The required point of intersection of three planes is:</p>
Step-by-Step Solution
Key Concept: To find the intersection point of three planes, solve the system of three linear equations simultaneously. The intersection of three non-parallel planes is a unique point (when they meet at one point).
Step 1: Eliminate x from equations (1) and (2):
(2) − (1): (x + 2y − 2z − 5) − (x − 2y + z − 1) = 0
4y − 3z − 4 = 0 ... (4) Step 2: Eliminate x from equations (1) and (3):
(3) − (1): (x + y − 2z − 7) − (x − 2y + z − 1) = 0
3y − 3z − 6 = 0
y − z − 2 = 0 → y = z + 2 ... (5) Step 3: Substitute (5) into (4):
4(z + 2) − 3z − 4 = 0
4z + 8 − 3z − 4 = 0
z + 4 = 0 → z = −4 Step 4: From (5): y = −4 + 2 = −2 Step 5: Substitute y = −2 and z = −4 into equation (1):
x − 2(−2) + (−4) − 1 = 0
x + 4 − 4 − 1 = 0
x = 1 Verification: Check (1,−2,−4) in all three equations:
(1): 1 − 2(−2) − 4 − 1 = 1 + 4 − 4 − 1 = 0 ✓
(2): 1 + 2(−2) − 2(−4) − 5 = 1 − 4 + 8 − 5 = 0 ✓
(3): 1 + (−2) − 2(−4) − 7 = 1 − 2 + 8 − 7 = 0 ✓ ∴ Answer: (1, −2, −4)
Correct Answer: A