The equation of the cone passing through the three coordinate axes and the lines \(\frac{x}{1} = \frac{y}{{ - 2}} = \frac{z}{3}\), \(\frac{x}{3} = \frac{y}{{ - 1}} = \frac{z}{1}\) is given by

Concept:

The general equation of a cone is given by

ax² + by² + cz² + 2fyz + 2gzx + 2hxy = 0

Where a, b, c, d, e, f are real numbers and a ≠ 0, b ≠ 0, c ≠ 0.

The point on three coordinate axes x, y, and z are

(x, 0, 0), (0, y, 0), and (0, 0, z).

Explanation:

The general equation of a cone with a vertex at the origin is given by

ax² + by² + cz² + 2fyz + 2gzx + 2hxy = 0    ----(1)

According to the question, the cone passes coordinate axes, it must satisfy 

(x, 0, 0), (0, y, 0), and (0, 0, z).

From here we will get that

a = 0, b = 0 and c = 0

Therefore, from equation (1)

2fyz + 2gzx + 2hxy = 0

⇒ fyz + gzx + hxy = 0      -----(2)

Given that the lines \(\frac{x}{1} = \frac{y}{{ - 2}} = \frac{z}{3}\) & \(\frac{x}{3} = \frac{y}{{ - 1}} = \frac{z}{1}\)  

are the generators of the cone, thus d.r.s of the lines satisfy the equation.

f(−2)(3) + g(3)(1) + h(1)(−2) = 0

⇒ −6f + 3g − 2h = 0       ------(3)

f(−1)(−1) + g(1)(3) + h(3)( −1) = 0

⇒ f − 3g − 3h = 0         -----(4)

\(⇒ \frac{f}{-9+6} =\frac{g}{2-18} = \frac{h}{-18+3} = k \)

⇒ f = −3k, g = −16k, h = -15k

From equation (3)

⇒ −3kyz − 16kzx - 15kxy = 0

∴  3yz + 16zx + 15xy = 0