Question

The volume of the tetrahedron whose vertices are A\((-1,2,3)\), B\((3,-2,1)\), C\((p,1,3)\), D\((-1,-2,4)\) is \(\frac{16}{3}\) cubic units then the value of p is

• A. \(\frac{-10}{3}\)
• B. \(5\)
• C. \(8\)
• D. \(10\)

Hint:

Volume = one sixth of the absolute scalar triple product of AB, AC, AD.

Answer 1

The Correct Option is A

Step 1: Plug the options:
Instead of solving, test $p = -\frac{10}{3}$: the triple product is $12\left(-\frac{10}{3}\right) + 8 = -40 + 8 = -32$.

Step 2: Volume:
$\frac16 \times 32 = \frac{16}{3}$, which matches the given volume.

Step 3: Check others:
$p = 5, 8, 10$ give triple products 68, 104, 128, so the volumes are $\frac{34}{3}$, $\frac{52}{3}$ and $\frac{64}{3}$, none equal to $\frac{16}{3}$.

Final Answer:
p equals -10/3, option (A). \[ \boxed{-\frac{10}{3}} \]