Equation of Y-axis  

Equation of Z-axis 

 None

---

Solution

Median of equilateral triangle = $\dfrac{\sqrt{3}}{2} \times \text{side}$ $\Rightarrow 3a = \dfrac{\sqrt{3}}{2} \times \text{side}$ $\Rightarrow \text{side} = 2\sqrt{3}a$ Radius (distance from centre to vertex) = $\dfrac{\text{side}}{\sqrt{3}} = 2a$. Equation: $x^2 + y^2 = (2a)^2 = 4a^2$. $\boxed{\text{Answer: (C) }x^2 + y^2 = 4a^2}$

---

Solution

Reflection in xy-plane → z-coordinate changes sign, x and y remain same. $\boxed{\text{Answer: (D) }(\alpha, \beta, -\gamma)}$

---

Solution

Given equation: $xy + yz = 0 \Rightarrow y(x + z) = 0$ This represents two planes: 1️⃣ $y = 0$ 2️⃣ $x + z = 0$ Normal to first plane = $(0,1,0)$ Normal to second plane = $(1,0,1)$ Their dot product = $0(1) + 1(0) + 0(1) = 0$, so the planes are perpendicular. $\boxed{\text{Answer: (D) A pair of perpendicular planes}}$

---

Solution

Distance from x-axis $= \sqrt{y^2 + z^2} = \sqrt{5^2 + 7^2} = \sqrt{74}$.

---

Solution

Distance equals $|x_2-x_1|$ when the segment is horizontal (same $y$), i.e., parallel to the $x$-axis.

---

Solution

Using triangle inequality,

$$|z_1 + z_2 + 3 + 4i| \le |z_1| + |z_2| + |3 + 4i| = 4 + 3 + 5 = 12.$$

Hence, value is less than 12.