1
MHT CET 2021 21th September Evening Shift
+2
-0

The equation of a line passing through $$(3,-1,2)$$ and perpendicular to the lines $$\bar{r}=(\hat{i}+\hat{j}-\hat{k})+\lambda(2 \hat{i}-2 \hat{j}+\hat{k})$$ and $$\bar{r}=(2 \hat{i}+\hat{j}-3 \hat{k})+\mu(\hat{i}-2 \hat{j}+2 \hat{k})$$ is

A
$$\frac{x-3}{2}=\frac{y+1}{3}=\frac{z-2}{2}$$
B
$$\frac{x-3}{3}=\frac{y+1}{2}=\frac{z-2}{2}$$
C
$$\frac{x+3}{2}=\frac{y+1}{3}=\frac{z-2}{2}$$
D
$$\frac{x-3}{2}=\frac{y+1}{2}=\frac{z-2}{3}$$
2
MHT CET 2021 21th September Evening Shift
+2
-0

The area of the parallelogram with vertices A(1, 2, 3), B(1, 3, a), C(3, 8, 6) and D(3, 7, 3) is $$\sqrt{265}$$ sq. units, then a =

A
$$-5,2$$
B
6
C
$$-6,0$$
D
6, 0
3
MHT CET 2021 21th September Morning Shift
+2
-0

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

A
$$\frac{-70}{11}$$
B
$$\frac{70}{11}$$
C
$$\frac{11}{70}$$
D
$$\frac{-11}{70}$$
4
MHT CET 2021 21th September Morning Shift
+2
-0

The Cartesian equation of the plane passing through the point A(7, 8, 6) and parallel to the XY plane is

A
z = 1
B
y = 8
C
x = 7
D
z = 6
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Physics
Mechanics
Optics
Electromagnetism
Modern Physics
Chemistry
Physical Chemistry
Inorganic Chemistry
Organic Chemistry
Mathematics
Algebra
Trigonometry
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Coordinate Geometry
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