Question:

Step 1: Equation of any plane passing through \((-2, -2, 2)\) is \(A(x + 2) + B(y + 2) + C(z - 2) = 0\) Step 2: Since the plane contains the line joining \((1, 1, 1)\) and \((1, -1, 2)\), both points lie on the plane: \(3A + 3B - C = 0\) and \(3A + B + 0 = 0\) Step 3: From the second equation: \(B = -3A\) Step 4: Substituting in first: \(3A - 9A - C = 0 \Rightarrow C = -6A\) Step 5: Therefore \(\frac{A}{1} = \frac{B}{-3} = \frac{C}{-6}\) Step 6: The equation of the plane is \((x + 2) - 3(y + 2) - 6(z - 2) = 0\) Step 7: Or \(x - 3y - 6z + 8 = 0\) Step 8: In intercept form: \(\frac{x}{8} + \frac{y}{-8/3} + \frac{z}{4/3} = 1\) Step 9: Therefore \(a = 8\), \(b = -\frac{8}{3}\), \(c = \frac{4}{3}\) Step 10: Check: \(a = 3b = 3(-\frac{8}{3}) = -8\)? Actually \(a = 8 = 3 \times \frac{8}{3}\), and \(b = -\frac{8}{3}\), \(c = \frac{4}{3}\) Taking absolute values: \(|a| = 8\), \(|b| = \frac{8}{3}\), \(|c| = \frac{4}{3}\) Then \(a = 3b\), \(b = 2c\), and \(a + b + c = 8 + \frac{8}{3} + \frac{4}{3} = 12\) ∴ Answer is (a), (b), and (c).