Question

If for some $\alpha ,\beta $ in $\mathbb{R}$, the intersection of the following three planes
$x + 4y - 2z = 1$
$x + 7y - 5z = \beta $
$x + 5y + \alpha z = 5$
Is a line in ${\mathbb{R}^3}$ then $\alpha + \beta $ is equal to?
A) $0$
B) $2$
C) $10$
D) $ - 10$

Answer 1

Hint:
We will use the formula to find the determinants corresponding to the intersection of the given planes. We will use the condition in which the intersection of planes is a line. We will obtain the unknowns and add them to find the answer of the given quantity.

Complete step by step solution:
The given equations are as follows:
$x + 4y - 2z = 1$
$x + 7y - 5z = \beta $
$x + 5y + \alpha z = 5$
Now consider the general equations of three planes as follows:
${a_1}x + {b_1}y + {c_1}z = {d_1}$
${a_2}x + {b_2}y + {c_2}z = {d_2}$
${a_3}x + {b_3}y + {c_3}z = {d_3}$
The we form the following determinants:
\[D = \left| {\begin{array}{*{20}{l}}
  {{a_1}}&{{b_1}}&{{c_1}} \\
  {{a_2}}&{{b_2}}&{{c_2}} \\
  {{a_3}}&{{b_3}}&{{c_3}}
\end{array}} \right|\]
Similarly, another determinant is given as follows:
\[{D_x} = \left| {\begin{array}{*{20}{l}}
  {{d_1}}&{{b_1}}&{{c_1}} \\
  {{d_2}}&{{b_2}}&{{c_2}} \\
  {{d_3}}&{{b_3}}&{{c_3}}
\end{array}} \right|\]
We form one more determinant as follows:
\[{D_y} = \left| {\begin{array}{*{20}{l}}
  {{a_1}}&{{d_1}}&{{c_1}} \\
  {{a_2}}&{{d_2}}&{{c_2}} \\
  {{a_3}}&{{d_3}}&{{c_3}}
\end{array}} \right|\]
And the final determinant is formed as follows:
\[{D_z} = \left| {\begin{array}{*{20}{l}}
  {{a_1}}&{{b_1}}&{{d_1}} \\
  {{a_2}}&{{b_2}}&{{d_2}} \\
  {{a_3}}&{{b_3}}&{{d_3}}
\end{array}} \right|\]
We say that the intersection of the planes is a line in ${\mathbb{R}^3}$ is a line if \[D = {D_x} = {D_y} = {D_z} = 0\] .
In the given equation it is given that the intersection of the planes is a line.
Therefore, the first condition will state that $D = 0$.
This will lead us to following:
$D = \left| {\begin{array}{*{20}{l}}
  1&4&{ - 2} \\
  1&7&{ - 5} \\
  1&5&\alpha
\end{array}} \right| \Rightarrow \left| {\begin{array}{*{20}{c}}
  1&4&{ - 2} \\
  1&7&{ - 5} \\
  1&5&\alpha
\end{array}} \right| = 0$
We will simplify the above determinant as follows:
$1\left( {7\alpha + 25} \right) - 4\left( {\alpha + 5} \right) - 2\left( {5 - 7} \right) = 0$
Simplify it further as follows:
$3\alpha + 9 = 0$
Therefore, solving for the unknown we get,
$\alpha = - 3$
similarly, another condition will state that ${D_z} = 0$.
This will lead us to following:
$D = \left| {\begin{array}{*{20}{l}}
  1&4&1 \\
  1&7&\beta \\
  1&5&5
\end{array}} \right| \Rightarrow \left| {\begin{array}{*{20}{c}}
  1&4&1 \\
  1&7&\beta \\
  1&5&5
\end{array}} \right| = 0$
We will simplify the above determinant as follows:
$1\left( {35 - 5\beta } \right) - 4\left( {5 - \beta } \right) + 1\left( {5 - 7} \right) = 0$
Simplify it further as follows:
$\beta - 13 = 0$
Therefore, solving for the unknown we get,
$\beta = 13$
Therefore, $\alpha + \beta = 10$

Hence, the correct option is C.

Note:
It is necessary to focus on the fact that the intersection of the planes is a line in the three-dimensional plane. Thus, we can use the condition where we equate the determinants to zero. It is also very important to perform all the calculations carefully.