Let $\bar{A B} = \hat{i} + \hat{j} - 3 \hat{k} . \bar{A C} = 3 \hat{i} + \hat{j} + 4 \hat{k} .$ and $\bar{A D} = 2 \hat{i} - \hat{k}$ are three co-terminus edges of a tetrahedron ABCD. If position vector of the centroid of tetrahedron is $(\hat{i} + 2 \hat{j} + 3 \hat{k})$ then

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Position vector of point A is $(2 \hat{i} + 3 \hat{k})$

Acute angle between the face ABC and ACD is $\cos^{- 1} \frac{146}{\sqrt{27972}}$

Length of perpendicular from D to the plane ABC is $\sqrt{\frac{128}{111}}$

Volume of tetrahedron is $\frac{8}{3}$

answer is B, C, D.

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Detailed Solution

Let $A (\vec{a}) \vec{c} (3 \hat{i} + \hat{j} + 4 \hat{k} + \vec{a})$
Now centre $= \hat{i} + 2 \hat{j} + 3 \hat{k} = \frac{4 \vec{a} + 6 \hat{i} + 2 \hat{j}}{4}$
   $\vec{a} = \frac{- \hat{i} + 3 \hat{j} + 6 \hat{k}}{2}$
Volume of tetrahedron
$= \frac{1}{6} [\bar{A B} \bar{A C} \bar{A D}] = \frac{8}{3}$
area of $Δ A B C = \frac{1}{2} | \bar{A B} \times \bar{A C} | = \sqrt{\frac{111}{2}}$
   $\frac{8}{3} = \frac{1}{3} (\sqrt{\frac{111}{2}}) \times h \Rightarrow h = \sqrt{\frac{128}{111}} \cos θ = \frac{(\bar{A B} . \bar{A C}) (\bar{A C} . \bar{A D})}{| \bar{A B} \times \bar{A C} | | \bar{A C} \times \bar{A D} |} \cos θ = \frac{(\bar{A B} . \bar{A C}) (\bar{A C} . \bar{A D}) - (\bar{A B} \times \bar{A D}) (\bar{A C} . \bar{A C})}{| \bar{A B} \times \bar{A C} | | \bar{A C} \times \bar{A D} |} \bar{A B} . \bar{A C} = - 8, \bar{A C} . \bar{A D} = 2 \bar{A B} . \bar{A D} = 5, \bar{A C} . \bar{A C} = 26 | \bar{A B} \times \bar{A C} | = \sqrt{222} \bar{A C} \times \bar{A D} = - \hat{i} + 11 \hat{j} - 2 \hat{k} | \bar{A C} \times \bar{A D} | = \sqrt{125 + 4 + 1} = \sqrt{126} \cos θ = \frac{146}{\sqrt{126} \sqrt{222}}$