Question

If the origin and point $P(2,3,4), Q(1,2,3)$ and $R(x,y,z)$ are coplanar then
(A) $x - 2y - z = 0$
(B) $x + 2y + z = 0$
(C) $x - 2y + z = 0$
(D) $2x - 2y + z = 0$

Answer 1

Hint: Let\[({x_1},{y_1},{z_1}) \equiv (2,3,4),({x_2},{y_2},{z_2}) \equiv (1,2,3),({x_3},{y_3},{z_3}) \equiv (x,y,z),({x_4},{y_4},{z_4}) \equiv (0,0,0)\]. Then solve the equation\[\left| {\begin{array}{*{20}{c}}
  {{x_4} - {x_1}}&{{y_4} - {y_1}}&{{z_4} - {z_1}} \\
  {{x_4} - {x_2}}&{{y_4} - {y_2}}&{{z_4} - {z_2}} \\
  {{x_4} - {x_3}}&{{y_4} - {y_3}}&{{z_4} - {z_3}}
\end{array}} \right| = 0\]. The equation of the plane thus obtained is the required answer.

Complete step by step answer:
 We are given the points $P(2,3,4)$, $Q(1,2,3)$ and $R(x,y,z)$and also the origin. These points are coplanar.
Let $O(0,0,0)$ be the origin of the plane containing these points.
Thus, O, P, Q, and R are co-planar points.
We know that if \[A({x_1},{y_1},{z_1}), B({x_2},{y_2},{z_2}), C({x_3},{y_3},{z_3}), D({x_4},{y_4},{z_4})\] are four points such that they are coplanar, then they satisfy the following condition:
\[\left| {\begin{array}{*{20}{c}}
  {{x_4} - {x_1}}&{{y_4} - {y_1}}&{{z_4} - {z_1}} \\
  {{x_4} - {x_2}}&{{y_4} - {y_2}}&{{z_4} - {z_2}} \\
  {{x_4} - {x_3}}&{{y_4} - {y_3}}&{{z_4} - {z_3}}
\end{array}} \right| = 0\]

Now, consider the points $O(0,0,0)$, $P(2,3,4)$, $Q(1,2,3)$ and $R(x,y,z)$.
Let
\[({x_1},{y_1},{z_1}) \equiv (2,3,4)\]
\[({x_2},{y_2},{z_2}) \equiv (1,2,3)\]
\[({x_3},{y_3},{z_3}) \equiv (x,y,z)\]
\[({x_4},{y_4},{z_4}) \equiv (0,0,0)\]
Then the equation \[\left| {\begin{array}{*{20}{c}}
  {{x_4} - {x_1}}&{{y_4} - {y_1}}&{{z_4} - {z_1}} \\
  {{x_4} - {x_2}}&{{y_4} - {y_2}}&{{z_4} - {z_2}} \\
  {{x_4} - {x_3}}&{{y_4} - {y_3}}&{{z_4} - {z_3}}
\end{array}} \right| = 0\]becomes
\[
  \left| {\begin{array}{*{20}{c}}
  {0 - 2}&{0 - 3}&{0 - 4} \\
  {0 - 1}&{0 - 2}&{0 - 3} \\
  {0 - x}&{0 - y}&{0 - z}
\end{array}} \right| = 0 \\
   \Rightarrow \left| {\begin{array}{*{20}{c}}
  { - 2}&{ - 3}&{ - 4} \\
  { - 1}&{ - 2}&{ - 3} \\
  { - x}&{ - y}&{ - z}
\end{array}} \right| = 0 \\
 \]
We can remove the negative signs inside the determinant in the LHS of the above equation by multiplying each row by $ - 1$.
Thus, we have
\[\left| {\begin{array}{*{20}{c}}
  2&3&4 \\
  1&2&3 \\
  x&y&z
\end{array}} \right| = 0....(1)\]
Solving the determinant in (1), we get
\[
  \left| {\begin{array}{*{20}{c}}
  2&3&4 \\
  1&2&3 \\
  x&y&z
\end{array}} \right| = 2(2z - 3y) - 3(z - 3x) + 4(y - 2x) \\
   = 4z - 6y - 3z + 9x + 4y - 8x \\
   = x - 2y + z........(2) \\
 \]
Using (1) and (2), we get the equation of the plane as \[x - 2y + z = 0\]

Hence \[x - 2y + z = 0\] is the correct answer.

Note: One of the properties of the determinant is as follows:
If we multiply any row or column of a determinant by a constant k, then the determinant gets multiplied by k.