Q74 / 150 · MHT CET 2026 (18 April Shift 1) · Mathematics
If the plane $2x + 3y + z = 6$ cuts coordinate axes at A, B and C, then the volume of tetrahedron OABC (where O is the origin) is ......cubic units.
- A$30$
- Correct answerB$6$
- C$36$
- D$180$
Solution
The equation of the plane is $2x + 3y + z = 6$.
Dividing by $6$, the intercept form of the plane is $\dfrac{x}{3} + \dfrac{y}{2} + \dfrac{z}{6} = 1$.
The intercepts on the coordinate axes are $a = 3$, $b = 2$, and $c = 6$.
The coordinates of the vertices of the tetrahedron are $A(3, 0, 0)$, $B(0, 2, 0)$, and $C(0, 0, 6)$, with $O(0, 0, 0)$ as the origin.
The volume of the tetrahedron $OABC$ is given by $V = \dfrac{1}{6}abc$.
Substituting the values, $V = \dfrac{1}{6}(3)(2)(6) = 6$ cubic units.
Answer: $6$
Dividing by $6$, the intercept form of the plane is $\dfrac{x}{3} + \dfrac{y}{2} + \dfrac{z}{6} = 1$.
The intercepts on the coordinate axes are $a = 3$, $b = 2$, and $c = 6$.
The coordinates of the vertices of the tetrahedron are $A(3, 0, 0)$, $B(0, 2, 0)$, and $C(0, 0, 6)$, with $O(0, 0, 0)$ as the origin.
The volume of the tetrahedron $OABC$ is given by $V = \dfrac{1}{6}abc$.
Substituting the values, $V = \dfrac{1}{6}(3)(2)(6) = 6$ cubic units.
Answer: $6$
Hence, the correct answer is option: 'B'.