Question:

Step 1: Find the direction vector of the line of intersection of the two planes $x - 2y + z - 1 = 0$ and $x + 2y - 2z - 5 = 0$. The normals are $\mathbf{n_1} = (1, -2, 1)$ and $\mathbf{n_2} = (1, 2, -2)$. $$\mathbf{d} = \mathbf{n_1} \times \mathbf{n_2} = \begin{vmatrix}\mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & -2 & 1 \\ 1 & 2 & -2\end{vmatrix} = \mathbf{i}(4-2) - \mathbf{j}(-2-1) + \mathbf{k}(2+2) = (2, 3, 4)$$ Step 2: Find a point on the line. Set $z = 0$: $x - 2y = 1$ and $x + 2y = 5$, giving $2x = 6 \Rightarrow x = 3, y = 1$. So a point on the line is $(3, 1, 0)$. Step 3: Parametric form of the line: $(x, y, z) = (3 + 2t, 1 + 3t, 4t)$. Step 4: Substitute into the plane $x + y - 2z = 7$: $$(3 + 2t) + (1 + 3t) - 2(4t) = 7$$ $$4 + 5t - 8t = 7 \Rightarrow 4 - 3t = 7 \Rightarrow t = -1$$ Step 5: Find the intersection point: $$\alpha = 3 + 2(-1) = 1, \quad \beta = 1 + 3(-1) = -2, \quad \gamma = 4(-1) = -4$$ So $(\alpha, \beta, \gamma) = (1, -2, -4)$. Step 6: Verify each option: - Option (a): $\alpha + \beta + \gamma = 1 - 2 - 4 = -5 \neq 7$. ✗ - Option (b): $\alpha - \beta + \gamma + 1 = 1 + 2 - 4 + 1 = 0$. ✓ - Option (c): $\alpha^2 + \beta^2 + \gamma^2 = 1 + 4 + 16 = 21$. ✓ - Option (d): $\alpha\beta + \beta\gamma + \gamma\alpha = (1)(-2) + (-2)(-4) + (-4)(1) = -2 + 8 - 4 = 2$. ✓ Step 7: The correct options are **(b), (c), and (d)**.