1
MHT CET 2023 13th May Evening Shift
+2
-0

The acute angle between the line joining the points $$(2,1,-3),(-3,1,7)$$ and a line parallel to $$\frac{x-1}{3}=\frac{y}{4}=\frac{z+3}{5}$$ through the point $$(-1,0,4)$$ is

A
$$\cos ^{-1}\left(\frac{1}{\sqrt{10}}\right)$$
B
$$\cos ^{-1}\left(\frac{5}{7 \sqrt{10}}\right)$$
C
$$\cos ^{-1}\left(\frac{7}{5 \sqrt{10}}\right)$$
D
$$\cos ^{-1}\left(\frac{3}{5 \sqrt{10}}\right)$$
2
MHT CET 2023 13th May Evening Shift
+2
-0

The foot of the perpendicular from the point $$(1,2,3)$$ on the line $$\mathbf{r}=(6 \hat{\mathbf{i}}+7 \hat{\mathbf{j}}+7 \hat{\mathbf{k}})+\lambda(3 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}-2 \hat{\mathbf{k}})$$ has the coordinates

A
$$(3,5,9)$$
B
$$(5,-3,9)$$
C
$$(3,-5,-9)$$
D
$$(5,-9,3)$$
3
MHT CET 2023 13th May Evening Shift
+2
-0

The distance of the point $$(1,6,2)$$ from the point of intersection of the line $$\frac{x-2}{3}=\frac{y+1}{4}=\frac{z-2}{12}$$ and the plane $$x-y+z=16$$ is

A
11 units
B
12 units
C
13 units
D
14 units
4
MHT CET 2023 13th May Morning Shift
+2
-0

A line drawn from the point $$\mathrm{A}(1,3,2)$$ parallel to the line $$\frac{x}{2}=\frac{y}{4}=\frac{z}{1}$$, intersects the plane $$3 x+y+2 z=5$$ in point $$\mathrm{B}$$, then co-ordinates of point $$\mathrm{B}$$ are

A
$$\left(\frac{1}{6}, \frac{4}{3}, \frac{19}{12}\right)$$
B
$$\left(-\frac{1}{6},-\frac{4}{3}, \frac{19}{12}\right)$$
C
$$\left(\frac{1}{6}, \frac{4}{3},-\frac{19}{12}\right)$$
D
$$\left(-\frac{1}{6},-\frac{4}{3},-\frac{19}{12}\right)$$
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