1
MHT CET 2021 23th September Morning Shift
+2
-0

If A(3, 2, $$-$$1), B($$-$$2, 2, $$-$$3) and D($$-$$2, 5, $$-$$4) are the vertices of a parallelogram, then the area of the parallelogram is

A
286 sq. units
B
$$\sqrt{286}$$ sq. units
C
300 sq. units
D
$$\sqrt{300}$$ sq. units
2
MHT CET 2021 23th September Morning Shift
+2
-0

The distance between the parallel lines $$\frac{x-2}{3}=\frac{y-4}{5}=\frac{z-1}{2}$$ and $$\frac{x-1}{3}=\frac{y+2}{5}=\frac{z+3}{2}$$ is

A
$$\frac{1}{\sqrt{38}}$$ units
B
$$\sqrt{\frac{333}{38}}$$ units
C
$$\sqrt{\frac{300}{37}}$$ units
D
$$\sqrt{\frac{300}{35}}$$ units
3
MHT CET 2021 23th September Morning Shift
+2
-0

The coordinates of the foot of the perpendicular drawn from the origin to the plane $$2 x+y-2 z=18$$ are

A
$$(4,2,-4)$$
B
$$(1,2,-3)$$
C
$$(4,2,4)$$
D
$$(4,-2,-4)$$
4
MHT CET 2021 23th September Morning Shift
+2
-0

The vector equation of the line passing through $$\mathrm{P}(1,2,3)$$ and $$\mathrm{Q}(2,3,4)$$ is

A
$$(\hat{i}+2 \hat{j}+3 \hat{k})+\lambda(\hat{i}+\hat{j}+\hat{k})$$
B
$$(\hat{i}+2 \hat{j}+3 \hat{k})+\lambda(\hat{i}-\hat{j}-\hat{k})$$
C
$$(\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}})+\lambda(2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+4 \hat{\mathrm{k}})$$
D
$$(\hat{i}+2 \hat{j}+3 \hat{k})+\lambda(2 \hat{i}+6 \hat{j}+12 \hat{k})$$
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