MHT CET 2024 2nd May Evening Shift | Three Dimensional Geometry Question 67 | Mathematics

Let $P$ be the image of the point $(3,1,7)$ with respect to the plane $x-y+z=3$. Then the equation of the plane passing through $P$ and containing the straight line $\frac{x}{1}=\frac{y}{2}=\frac{z}{1}$ is

$-4 y-x+7 z=0$

$x-4 y-7 z=0$

$x-4 y+7 z=0$

$x+4 y+7 z=0$

MHT CET 2024 2nd May Evening Shift

MCQ (Single Correct Answer)

-0

The incentre of the triangle whose vertices are $P(0,3,0), Q(0,0,4)$ and $R(0,3,4)$ is

$(0,3,2)$

$(0,2,3)$

$(2,0,3)$

$(2,3,0)$

MHT CET 2024 2nd May Morning Shift

MCQ (Single Correct Answer)

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The vector equation of a line whose Cartesian equations are $y=2,4 x-3 z+5=0$ is

$\overline{\mathrm{r}}=(3 \hat{\mathrm{i}}+4 \mathrm{k})+\lambda\left(2 \hat{\mathrm{j}}+\frac{5}{3} \hat{\mathrm{k}}\right)$

$\overline{\mathrm{r}}=(3 \hat{\mathrm{i}}+4 \mathrm{k})+\lambda\left(2 \hat{\mathrm{j}}-\frac{5}{3} \hat{\mathrm{k}}\right)$

$\overline{\mathrm{r}}=\left(2 \hat{\mathrm{j}}+\frac{5}{3} \hat{\mathrm{k}}\right)+\lambda(3 \hat{\mathrm{i}}+4 \mathrm{k})$

$\overline{\mathrm{r}}=\left(2 \hat{\mathrm{j}}-\frac{5}{3} \hat{\mathrm{k}}\right)+\lambda(3 \hat{\mathrm{i}}+4 \mathrm{k})$

MHT CET 2024 2nd May Morning Shift

MCQ (Single Correct Answer)

-0

The Cartesian equation of the plane, passing through the points $(3,1,1),(1,2,3)$ and $(-1,4,2)$, is

$5 x+6 y-2 z-23=0$

$-5 x+6 y+2 z+23=0$

$5 x+6 y+2 z-23=0$

$5 x-6 y+2 z-23=0$

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