CUET UG 2025 — Mathematics Geometry

The line equation is $\frac{x+3}{3} = \frac{y-1}{-1} = \frac{z-5}{-5}$3x+3​=−1y−1​=−5z−5​

When a line cuts the yz-plane, the x-coordinate is 0 at that point. The yz-plane is where $x = 0$x=0.

---

Since all three fractions are equal, let this common value be $λ$λ.

$$\frac{x+3}{3} = λ$$3x+3​=λ

$$\frac{y-1}{-1} = λ$$−1y−1​=λ

$$\frac{z-5}{-5} = λ$$−5z−5​=λ

---

From the first equation:

$$\frac{x+3}{3} = λ$$3x+3​=λ

$$x + 3 = 3λ$$x+3=3λ

Since the line cuts the yz-plane, $x = 0$x=0:

$$0 + 3 = 3λ$$0+3=3λ

$$λ = 1$$λ=1

---

From the second equation with $λ = 1$λ=1:

$$\frac{y-1}{-1} = 1$$−1y−1​=1

$$y - 1 = -1$$y−1=−1

$$y = 0$$y=0

---

From the third equation with $λ = 1$λ=1:

$$\frac{z-5}{-5} = 1$$−5z−5​=1

$$z - 5 = -5$$z−5=−5

$$z = 0$$z=0

---

Therefore, the coordinates are $(0, 0, 0)$(0,0,0).