MHT CET 2023 12th May Morning Shift | Three Dimensional Geometry Question 31 | Mathematics

If the line $$\frac{1-x}{3}=\frac{7 y-14}{2 p}=\frac{z-3}{2}$$ and $$\frac{7-7 x}{3 \mathrm{p}}=\frac{y-5}{1}=\frac{6-\mathrm{z}}{5}$$ are at right angles, then $$\mathrm{p}=$$

$$\frac{70}{11}$$

$$\frac{11}{70}$$

$$\frac{-70}{11}$$

$$\frac{-11}{70}$$

MHT CET 2023 11th May Evening Shift

MCQ (Single Correct Answer)

-0

If the lines $\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-1}{4}$ and $x-3=\frac{y-\mathrm{k}}{2}=\mathrm{z}$ intersect, then the value of $\mathrm{k}$ is

$$\frac{3}{2}$$

$$\frac{-2}{9}$$

$$\frac{-2}{3}$$

$$\frac{9}{2}$$

MHT CET 2023 11th May Evening Shift

MCQ (Single Correct Answer)

-0

If the line $$\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-2}{4}$$ meets the plane $$x+2 y+3 z=15$$ at the point $$P$$, then the distance of $$\mathrm{P}$$ from the origin is

$$\frac{7}{2}$$ units

$$\frac{9}{2}$$ units

$$\frac{\sqrt{5}}{2}$$ units

$$2 \sqrt{5}$$ units

MHT CET 2023 11th May Evening Shift

MCQ (Single Correct Answer)

-0

The equation of line passing through the point $$(1,2,3)$$ and perpendicular to the lines $$\frac{x-2}{3}=\frac{y-1}{2}=\frac{z+1}{-2}$$ and $$\frac{x}{2}=\frac{y}{-3}=\frac{z}{1}$$ is

$$\overline{\mathrm{r}}=(\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}})+\lambda(4 \hat{\mathrm{i}}+7 \hat{\mathrm{j}}-13 \hat{\mathrm{k}})$$

$$\overline{\mathrm{r}}=(\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}})+\lambda(-4 \hat{\mathrm{i}}+7 \hat{\mathrm{j}}-13 \hat{\mathrm{k}})$$

$$\overline{\mathrm{r}}=(\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}})+\lambda(-4 \hat{\mathrm{i}}-7 \hat{\mathrm{j}}-13 \hat{\mathrm{k}})$$

$$\overline{\mathrm{r}}=(\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}})+\lambda(4 \hat{\mathrm{i}}-7 \hat{\mathrm{j}}-13 \hat{\mathrm{k}})$$

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