Question:
If for some \(\alpha\) and \(\beta\) in R, the intersection of the following three places
\(x + 4y - 2z = 1\)
\(x + 7y - 5z = b\)
\(x + 5y + \alpha z = 5\)
is a line in \(\mathbb{R}^3\), then \(\alpha + \beta\) is equal to :
If for some \(\alpha\) and \(\beta\) in R, the intersection of the following three places
\(x + 4y - 2z = 1\)
\(x + 7y - 5z = b\)
\(x + 5y + \alpha z = 5\)
is a line in \(\mathbb{R}^3\), then \(\alpha + \beta\) is equal to :
\(x + 4y - 2z = 1\)
\(x + 7y - 5z = b\)
\(x + 5y + \alpha z = 5\)
is a line in \(\mathbb{R}^3\), then \(\alpha + \beta\) is equal to :
Show Hint
Three planes meet in a line only if the coefficient determinant is zero and the system is consistent (infinitely many solutions). Treat the constant b as \(\beta\).
Updated On: Nov 5, 2026
- -10
- 0
- 10
- 2
Show Solution
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The Correct Option is C
Solution and Explanation
The printed question says 'places' (planes) and the constant is written b, but the options ask for \(\alpha+\beta\); b is taken as \(\beta\).
Subtract equation 1 from equations 2 and 3: \(3y - 3z = b-1\) and \(y + (\alpha+2)z = 4\).
For infinitely many solutions these two equations must be dependent: \(\frac{1}{3} = \frac{\alpha+2}{-3} = \frac{4}{b-1}\).
From \(\frac{1}{3} = \frac{\alpha+2}{-3}\): \(\alpha + 2 = -1\), so \(\alpha = -3\).
From \(\frac{1}{3} = \frac{4}{b-1}\): \(b - 1 = 12\), so \(b = \beta = 13\).
\(\alpha + \beta = -3 + 13 = 10\).
Answer: (3) 10
Subtract equation 1 from equations 2 and 3: \(3y - 3z = b-1\) and \(y + (\alpha+2)z = 4\).
For infinitely many solutions these two equations must be dependent: \(\frac{1}{3} = \frac{\alpha+2}{-3} = \frac{4}{b-1}\).
From \(\frac{1}{3} = \frac{\alpha+2}{-3}\): \(\alpha + 2 = -1\), so \(\alpha = -3\).
From \(\frac{1}{3} = \frac{4}{b-1}\): \(b - 1 = 12\), so \(b = \beta = 13\).
\(\alpha + \beta = -3 + 13 = 10\).
Answer: (3) 10
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