1
MHT CET 2023 11th May Evening Shift
MCQ (Single Correct Answer)
+2
-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

A
$$\frac{7}{2}$$ units
B
$$\frac{9}{2}$$ units
C
$$\frac{\sqrt{5}}{2}$$ units
D
$$2 \sqrt{5}$$ units
2
MHT CET 2023 11th May Evening Shift
MCQ (Single Correct Answer)
+2
-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

A
$$\overline{\mathrm{r}}=(\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}})+\lambda(4 \hat{\mathrm{i}}+7 \hat{\mathrm{j}}-13 \hat{\mathrm{k}})$$
B
$$\overline{\mathrm{r}}=(\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}})+\lambda(-4 \hat{\mathrm{i}}+7 \hat{\mathrm{j}}-13 \hat{\mathrm{k}})$$
C
$$\overline{\mathrm{r}}=(\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}})+\lambda(-4 \hat{\mathrm{i}}-7 \hat{\mathrm{j}}-13 \hat{\mathrm{k}})$$
D
$$\overline{\mathrm{r}}=(\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}})+\lambda(4 \hat{\mathrm{i}}-7 \hat{\mathrm{j}}-13 \hat{\mathrm{k}})$$
3
MHT CET 2023 11th May Evening Shift
MCQ (Single Correct Answer)
+2
-0

The angle between the line $$\frac{x+1}{2}=\frac{y-2}{1}=\frac{z-3}{-2}$$ and plane $$x-2 y-\lambda z=3$$ is $$\cos ^{-1}\left(\frac{2 \sqrt{2}}{3}\right)$$, then value of $$\lambda$$ is

A
$$\sqrt{\frac{3}{5}}$$
B
$$\frac{5}{\sqrt{3}}$$
C
$$\sqrt{\frac{5}{3}}$$
D
$$\frac{1}{\sqrt{3}}$$
4
MHT CET 2023 11th May Morning Shift
MCQ (Single Correct Answer)
+2
-0

If the direction cosines $$l, \mathrm{~m}, \mathrm{n}$$ of two lines are connected by relations $$l-5 \mathrm{~m}+3 \mathrm{n}=0$$ and $$7 l^2+5 \mathrm{~m}^2-3 \mathrm{n}^2=0$$, then value of $$l+\mathrm{m}+\mathrm{n}$$ is

A
$$\frac{2}{\sqrt{6}}$$ or $$\frac{6}{\sqrt{14}}$$
B
$$\frac{1}{\sqrt{6}}$$ or $$\frac{5}{\sqrt{14}}$$
C
$$\frac{2}{\sqrt{6}}$$ or $$\frac{5}{\sqrt{14}}$$
D
$$\frac{1}{\sqrt{6}}$$ or $$\frac{6}{\sqrt{14}}$$
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