MHT CET 2026 19th April Morning Shift | Three Dimensional Geometry Question 15 | Mathematics

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

$\dfrac{x}{a} - \dfrac{y}{b} - \dfrac{z}{c} = 0$

$\dfrac{x}{a} - \dfrac{y}{b} + \dfrac{z}{c} = 0$

$\dfrac{x}{a} + \dfrac{y}{b} - \dfrac{z}{c} = 0$

$\dfrac{x}{a} + \dfrac{y}{b} + \dfrac{z}{c} = 0$

MHT CET 2026 19th April Morning Shift

MCQ (Single Correct Answer)

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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.)

$5$

$20$

$10$

$8$

MHT CET 2026 19th April Morning Shift

MCQ (Single Correct Answer)

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The symmetric form of the equation of the line $x = ay + b$, $z = cy + d$ is

$\dfrac{x - a}{b} = y = \dfrac{z - c}{d}$

$\dfrac{x - b}{a} = \dfrac{y - d}{c} = z$

$x = \dfrac{y - a}{b} = \dfrac{z - c}{d}$

$\dfrac{x - b}{a} = y = \dfrac{z - d}{c}$

MHT CET 2026 18th April Evening Shift

MCQ (Single Correct Answer)

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The equation of the plane passing through the points having position vectors $(\bar{a} + \bar{b}), (\bar{b} + \bar{c})$ and $(\bar{a} + \bar{c})$ is $\ldots$

$\bar{r} \cdot (\bar{a} \times \bar{b} + \bar{b} \times \bar{c} + \bar{c} \times \bar{a}) = [\bar{a}\ \bar{b}\ \bar{c}]$

$\bar{r} \cdot (\bar{a} \times \bar{b} + \bar{b} \times \bar{c} + \bar{c} \times \bar{a}) = 2[\bar{a}\ \bar{b}\ \bar{c}]$

$\bar{r} \cdot (\bar{a} \times \bar{b} + \bar{b} \times \bar{c} + \bar{a} \times \bar{c}) = [\bar{a}\ \bar{b}\ \bar{c}]$

$\bar{r} \cdot (\bar{a} \times \bar{b} + \bar{b} \times \bar{c} + \bar{a} \times \bar{c}) = 2[\bar{a}\ \bar{b}\ \bar{c}]$

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