MHT CET 2025 5th May Evening Shift | Three Dimensional Geometry Question 5 | Mathematics

A line L is passing through points $\mathrm{A}(1,3,2)$ and $\mathrm{B}(2,2,1)$. If mirror image of point $\mathrm{P}(1,1,-1)$ in the line L is $(x, y, z)$ then $x+y+\mathrm{z}=$

$\frac{10}{3}$

$\frac{13}{3}$

$\frac{14}{3}$

$\frac{23}{3}$

MHT CET 2025 26th April Evening Shift

MCQ (Single Correct Answer)

-0

The lines $\frac{6 x-6}{18}=\frac{y+1}{3}=\frac{z-1}{5} \quad$ and $\frac{3 x+6}{12}=\frac{y-1}{3}=\frac{z+1}{2}$ are $\ldots$

intersecting at point $(1,-1,2)$

intersecting at right angles

do not intersect

intersecting at point $(3,1,-1)$

MHT CET 2025 26th April Evening Shift

MCQ (Single Correct Answer)

-0

The line $\frac{x-1}{2}=\frac{y+2}{-1}=\frac{z}{1}$ intersects the XY plane and the YZ plane at points A and B respectively. The equation of line through the points A and B is

$[\overline{\mathrm{r}}-(\hat{\mathrm{i}}-2 \hat{\mathrm{j}}+0 \hat{\mathrm{k}})] \times\left(-\hat{\mathrm{i}}+\frac{1}{2} \hat{\mathrm{j}}-\frac{1}{2} \hat{\mathrm{k}}\right)=\overline{0}$

$[\overline{\mathrm{r}}+(\hat{\mathrm{i}}-2 \hat{\mathrm{j}}+0 \hat{\mathrm{k}})] \times\left(-\hat{\mathrm{i}}+\frac{1}{2} \hat{\mathrm{j}}+\frac{1}{2} \hat{\mathrm{k}}\right)=\overline{0}$

$\overline{\mathrm{r}}=(-\hat{\mathrm{i}}-2 \hat{\mathrm{j}}+0 \hat{\mathrm{k}})+\lambda\left(-\hat{\mathrm{i}}+\frac{1}{2} \hat{\mathrm{j}}-\frac{1}{2} \hat{\mathrm{k}}\right)$

$\quad \overline{\mathrm{r}}=(\hat{\mathrm{i}}+2 \hat{\mathrm{j}})+\lambda\left(-\hat{\mathrm{i}}+\frac{1}{2} \hat{\mathrm{j}}-\frac{1}{2} \hat{\mathrm{k}}\right)$

MHT CET 2025 26th April Evening Shift

MCQ (Single Correct Answer)

-0

The distance of the point $\mathrm{A}(3,-4,5)$ from the plane $2 x+5 y-6 z=16$ measured along the line $\frac{x}{2}=\frac{y}{1}=\frac{z}{-2}$ is

$\frac{60}{7}$ units

$\frac{7}{60}$ units

$\frac{1}{7}$ units

$\frac{2}{7}$ units

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