Question

If the origin and point $P\left( {2,3,4} \right),Q\left( {1,2,3} \right)$ and $R\left( {x,y,z} \right)$ 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:
It is given in the question that the origin and point $P\left( {2,3,4} \right),Q\left( {1,2,3} \right)$ and $R\left( {x,y,z} \right)$ are coplanar then what is the equation:
Firstly, we will assume $O\left( {0,0,0} \right)$ be the origin of the plane containing point P, Q and R.
Then after, if the points are coplanar then the equation satisfy is $\left| {\begin{array}{*{20}{l}}
  {{x_4} - {x_1}}&{{y_4} - {y_1}}&{{z_4} - {z_1}} \\
  {{x_4} - {x_2}}&{{y_4} - {y_2}}&{{z_4} - {z_2}} \\
  {{x_4} - {x_3}}&{{y_4} - {y_3}}&{{z_4} - {z_3}}
\end{array}} \right| = 0$ .
Next, we will put the values of P, Q, R and O in the above equation, and Finally after solving the determinant we will get our answer.

Complete step by step solution:
It is given in the question that the origin and point $P\left( {2,3,4} \right),Q\left( {1,2,3} \right)$ and $R\left( {x,y,z} \right)$ are coplanar then what is the equation:
Let $O\left( {0,0,0} \right)$ be the origin of the plane containing point P, Q and R.
Thus, O, P, Q and R are collinear points.
We know that, if $A\left( {{x_1},{y_1},{z_1}} \right),B\left( {{x_2},{y_2},{z_2}} \right),C\left( {{x_3},{y_3},{z_3}} \right)$ and $D\left( {{x_4},{y_4},{z_4}} \right)$ are four points such that they are coplanar, then they satisfy the following condition:
 $\left| {\begin{array}{*{20}{l}}
  {{x_4} - {x_1}}&{{y_4} - {y_1}}&{{z_4} - {z_1}} \\
  {{x_4} - {x_2}}&{{y_4} - {y_2}}&{{z_4} - {z_2}} \\
  {{x_4} - {x_3}}&{{y_4} - {y_3}}&{{z_4} - {z_3}}
\end{array}} \right| = 0$
Now, consider the points $O\left( {0,0,0} \right),$ $P\left( {2,3,4} \right),Q\left( {1,2,3} \right)$ and $R\left( {x,y,z} \right)$
Let,
 $
  \left( {{x_1},{y_1},{z_1}} \right) = \left( {2,3,4} \right) \\
  \left( {{x_2},{y_2},{z_2}} \right) = \left( {1,2,3} \right) \\
  \left( {{x_3},{y_3},{z_3}} \right) = \left( {x,y,z} \right) \\
  \left( {{x_4},{y_4},{z_4}} \right) = \left( {0,0,0} \right) \\
 $
Now, put the value of $\left( {{x_1},{y_1},{z_1}} \right),\left( {{x_2},{y_2},{z_2}} \right),\left( {{x_3},{y_3},{z_3}} \right),\left( {{x_4},{y_4},{z_4}} \right)$ in above equation.
Then the equation $\left| {\begin{array}{*{20}{l}}
  {{x_4} - {x_1}}&{{y_4} - {y_1}}&{{z_4} - {z_1}} \\
  {{x_4} - {x_2}}&{{y_4} - {y_2}}&{{z_4} - {z_2}} \\
  {{x_4} - {x_3}}&{{y_4} - {y_3}}&{{z_4} - {z_3}}
\end{array}} \right| = 0$ become \[\left| {\begin{array}{*{20}{l}}
  {0 - 2}&{0 - 3}&{0 - 4} \\
  {0 - 1}&{0 - 2}&{0 - 3} \\
  {0 - x}&{0 - y}&{0 - z}
\end{array}} \right| = 0\]
 $\therefore \left| {\begin{array}{*{20}{l}}
  { - 2}&{ - 3}&{ - 4} \\
  { - 1}&{ - 2}&{ - 3} \\
  { - x}&{ - y}&{ - z}
\end{array}} \right| = 0$
Now, we can remove the negative sign inside the determinant in the L.H.S of the above equation by multiplying each row by -1.
Thus, we have
 \[\left| {\begin{array}{*{20}{l}}
  2&3&4 \\
  1&2&3 \\
  x&y&z
\end{array}} \right| = 0\] (I)
Now, by solving the above determinant, we get,
 \[\therefore \left| {\begin{array}{*{20}{l}}
  2&3&4 \\
  1&2&3 \\
  x&y&z
\end{array}} \right| = 2\left( {2z - 3y} \right) - 3\left( {z - 3x} \right) + 4\left( {y - 2x} \right)\]
 $ = 4z - 6y - 3z + 9x + 4y - 8x$
 $ = x - 2y + z$ (II)
Now, by using equation (I) and (II), we get the equation of the plane as $x - 2y + z$ .

Hence, the equation of the plane is $x - 2y + z$.

Note:
One of the properties of the determinant is as follows:
Since, if we multiply any row or column of a determinant by a constant k, then the determinant gets multiplied by k.
Determinant: The determinant is a scalar value that can be computed from the elements of a square matrix and encodes certain properties of the linear transformation described by the matrix. The determinant of a matrix A is denoted as $\det \left( A \right),\det A$ or $\left| A \right|$.