Question

If for some $ \alpha $ and $ \beta $ in 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 $ {R^3} $ , the $ \alpha + \beta $ is equal to ?
A. 0
B. 2
C. 10
D. –10

Answer 1

Hint: In this question 3 planes are given whose interaction is a line.
And we know that, when the three planes share either no points, or else infinitely many points in either a line or a plane of intersection their determinant should be equal to zero. And determinant is denoted by D.
Here in this question we will use this concept to make equations to find the values of unknowns.Since we have two unknowns i.e., $ \alpha $ and $ \beta $ we will make two equations.Now we will discuss about determinant,
Since we will deal with $ 3 \times 3 $ matrix we will learn formula for $ 3 \times 3 $
Matrix :
 $ \left| {\begin{array}{*{20}{c}}
  a&b&c \\
  d&e&f \\
  g&h&i
\end{array}} \right| = a(ei - fh) - b(di - fg) + c(dh - eg) $

Complete step-by-step answer:
In the question we have given that :
\[x + 4y - 2z = 1\]
\[x + 7y - 5z = \beta \]
\[x + 5y + \alpha z = 5\]
It is also given that the given planes intersect in a line, so we can write,
\[D = Dx = Dy = Dz = 0\]
Or, we can say that, \[D = 0\]
As explained in the hint, D is determinant here.
Now, we will use this equation i.e., $ D = 0 $ to find unknowns in above equations.
 $ \left| {\begin{array}{*{20}{c}}
  1&4&{ - 2} \\
  1&7&{ - 5} \\
  1&5&\alpha
\end{array}} \right| = 0 $
Here we will solve the determinant D to get value of $ \alpha $
\[ \Rightarrow 1(7\alpha + 25)-4(\alpha + 5) + ( - 2)(5 - 7) = 0\]
\[ \Rightarrow 7\alpha + 25 - 4\alpha - 20 - 10 + 14 = 0\]
 $ \Rightarrow 3\alpha + 9 = 0 $
 $ \Rightarrow \alpha = - 3 $
And also \[{D_z} = 0\]
We will use the equation for \[{D_z} = 0\] to get the unknowns i.e., $ \beta $ here.
 $ \left| {\begin{array}{*{20}{c}}
  1&4&1 \\
  1&7&\beta \\
  1&5&5
\end{array}} \right| = 0 $
Here we will solve the determinant, $ {D_z} $ to get value of $ \beta $
\[ \Rightarrow 1(35 - 5\beta )-4(5 - \beta ) + 1(5 - 7) = 0\]
\[ \Rightarrow 35 - 5\beta - 20 + 4\beta + 5-7 = 0\]
 $ \Rightarrow 13 - \beta = 0 $
 $ \Rightarrow \beta = 13 $
Now we have both the values $ \alpha = - 3 $ and $ \beta = 13 $
According to question we need to find $ \alpha + \beta $
So, by adding both the values obtained from our calculation we got our final result
 $ \alpha + \beta = - 3 + 13 = 10 $
So, the correct answer is “Option C”.

Note: 1. The determinant is positive or negative according to whether it is linear.
2. It is to be noted that while finding determinants 2nd term should be taken as negative.
(i.e., Respective of j)