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

The co-ordinates of the point $$\mathrm{P} \equiv(1,2,3)$$ and $$\mathrm{O} \equiv(0,0,0)$$, then the direction cosines of $$\overline{\mathrm{OP}}$$ are

A
$$\frac{1}{\sqrt{14}}, \frac{2}{\sqrt{14}}, \frac{3}{\sqrt{14}}$$
B
$$\frac{1}{\sqrt{6}}, \frac{2}{\sqrt{6}}, \frac{1}{\sqrt{6}}$$
C
$$\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}$$
D
$$\frac{2}{\sqrt{29}}, \frac{3}{\sqrt{29}}, \frac{4}{\sqrt{29}}$$
2
MHT CET 2021 21th September Evening Shift
+2
-0

The equation of the plane containing the line $$\frac{x+1}{-3}=\frac{y-3}{2}=\frac{z+2}{1}$$ and the point $$(0,7,-7)$$ is

A
$$2 x+y+z=0$$
B
$$x+y+z=0$$
C
$$x+2 y-3 z=35$$
D
$$x+3 y+z=14$$
3
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
$$\frac{x-3}{2}=\frac{y+1}{2}=\frac{z-2}{3}$$
4
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
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