1
MHT CET 2026 19th April Morning Shift
MCQ (Single Correct Answer)
+2
-0
The plane $\dfrac{x}{2} + \dfrac{y}{3} + \dfrac{z}{4} = 1$ cuts the axes at the points A, B, C then the area of triangle ABC is
A
$\sqrt{29}$
B
$\sqrt{41}$
C
$\sqrt{61}$
D
$\sqrt{51}$
2
MHT CET 2026 19th April Morning Shift
MCQ (Single Correct Answer)
+2
-0
From a point P $(a, b, c)$, perpendiculars PA and PB are drawn to XY plane and ZX plane respectively. If O is the origin, then the equation of plane OAB is
A
$\dfrac{x}{a} - \dfrac{y}{b} - \dfrac{z}{c} = 0$
B
$\dfrac{x}{a} - \dfrac{y}{b} + \dfrac{z}{c} = 0$
C
$\dfrac{x}{a} + \dfrac{y}{b} - \dfrac{z}{c} = 0$
D
$\dfrac{x}{a} + \dfrac{y}{b} + \dfrac{z}{c} = 0$
3
MHT CET 2026 19th April Morning Shift
MCQ (Single Correct Answer)
+2
-0
If $p$ is the shortest distance between the lines $\dfrac{x+1}{7} = \dfrac{y+1}{-6} = z+1$ and $\vec{r} = (3\hat{i} + 5\hat{j} + 7\hat{k}) + \mu(\hat{i} - 2\hat{j} + \hat{k})$ then $[p]$ is... ,(where $[\,.\,]$ denotes the greatest integer function.)
A
$5$
B
$20$
C
$10$
D
$8$
4
MHT CET 2026 19th April Morning Shift
MCQ (Single Correct Answer)
+2
-0
The symmetric form of the equation of the line $x = ay + b$, $z = cy + d$ is
A
$\dfrac{x - a}{b} = y = \dfrac{z - c}{d}$
B
$\dfrac{x - b}{a} = \dfrac{y - d}{c} = z$
C
$x = \dfrac{y - a}{b} = \dfrac{z - c}{d}$
D
$\dfrac{x - b}{a} = y = \dfrac{z - d}{c}$

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