1
MHT CET 2023 9th May Evening Shift
+2
-0

The equation of a plane, containing the line of intersection of the planes $$2 x-y-4=0$$ and $$y+2 z-4=0$$ and passing through the point $$(2,1,0)$$, is

A
$$3 x-2 y+z=4$$
B
$$3 x+2 y+z=4$$
C
$$3 x-2 y-z=4$$
D
$$3 x+2 y-z=-4$$
2
MHT CET 2023 9th May Morning Shift
+2
-0

The foot of the perpendicular drawn from the origin to the plane is $$(4,-2,5)$$, then the Cartesian equation of the plane is

A
$$4 x-2 y+5 z=45$$
B
$$-4 x+2 y+5 z=45$$
C
$$4 x-2 y+5 z+45=0$$
D
$$4 x+2 y-5 z+45=0$$
3
MHT CET 2023 9th May Morning Shift
+2
-0

A vector $$\overrightarrow{\mathrm{n}}$$ is inclined to $$\mathrm{X}$$-axis at $$45^{\circ}$$, $$\mathrm{Y}$$-axis at $$60^{\circ}$$ and at an acute angle to Z-axis If $$\overrightarrow{\mathrm{n}}$$ is normal to a plane passing through the point $$(-\sqrt{2}, 1,1)$$, then equation of the plane is

A
$$\sqrt{2} x+y+z=0$$
B
$$x+\sqrt{2} y+z=1$$
C
$$-\sqrt{2} x+y+2 z=5$$
D
$$x+y+\sqrt{2} z=1$$
4
MHT CET 2023 9th May Morning Shift
+2
-0

If the Cartesian equation of a line is $$6 x-2=3 y+1=2 z-2$$, then the vector equation of the line is

A
$$\overline{\mathrm{r}}=\left(\frac{1}{3} \hat{\mathrm{i}}-\frac{1}{3} \hat{\mathrm{j}}+\hat{\mathrm{k}}\right)+\lambda(\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}})$$
B
$$\overline{\mathrm{r}}=(\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}})+\lambda(\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}})$$
C
$$\overline{\mathrm{r}}=\left(\frac{-1}{3} \hat{\mathrm{i}}+\frac{1}{3} \hat{\mathrm{j}}+\hat{\mathrm{k}}\right)+\lambda(\hat{\mathrm{i}}-2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}})$$
D
$$\overline{\mathrm{r}}=\left(\frac{1}{3} \hat{\mathrm{i}}-\frac{1}{3} \hat{\mathrm{j}}-\hat{\mathrm{k}}\right)+\lambda(\hat{\mathrm{i}}-\hat{\mathrm{j}}+\hat{\mathrm{k}})$$
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