1
MHT CET 2023 11th May Morning Shift
+2
-0

The vector equation of the line $$2 x+4=3 y+1=6 z-3$$ is

A
$$\overline{\mathrm{r}}=\left(2 \hat{\mathrm{i}}+\frac{1}{3} \hat{\mathrm{j}}+\frac{1}{2} \hat{\mathrm{k}}\right)+\lambda(3 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}+\overline{\mathrm{k}})$$
B
$$\overline{\mathrm{r}}=\left(-2 \hat{\mathrm{i}}-\frac{1}{3} \hat{\mathrm{j}}+\frac{1}{2} \hat{\mathrm{k}}\right)+\lambda(3 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}+\hat{\mathrm{k}})$$
C
$$\overline{\mathrm{r}}=(2 \hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}})+\lambda(3 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}+\overline{\mathrm{k}})$$
D
$$\overline{\mathrm{r}}=(-2 \hat{\mathrm{i}}-\hat{\mathrm{j}}+\hat{\mathrm{k}})+\lambda(3 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}+\hat{\mathrm{k}})$$
2
MHT CET 2023 10th May Evening Shift
+2
-0

The plane through the intersection of planes $$x+y+z=1$$ and $$2 x+3 y-z+4=0$$ and parallel to $$\mathrm{Y}$$-axis also passes through the point

A
$$(3,3,-1)$$
B
$$(-3,0,1)$$
C
$$(3,2,1)$$
D
$$(-3,0,-1)$$
3
MHT CET 2023 10th May Evening Shift
+2
-0

The perpendicular distance of the origin from the plane $$x-3 y+4 z-6=0$$ is

A
6
B
$$\frac{6}{\sqrt{26}}$$
C
$$\frac{1}{\sqrt{26}}$$
D
$$\frac{3}{\sqrt{26}}$$
4
MHT CET 2023 10th May Evening Shift
+2
-0

Two lines $$\frac{x-3}{1}=\frac{y+1}{3}=\frac{z-6}{-1}$$ and $$\frac{x+5}{7}=\frac{y-2}{-6}=\frac{z-3}{4} \quad$$ intersect at the point R. Then reflection of $$\mathrm{R}$$ in the $$x y$$-plane has co-ordinates

A
$$(2,-4,-7)$$
B
$$(2,-4,7)$$
C
$$(-2,4,7)$$
D
$$(2,4,7)$$
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