Mathematics · Straight lines

JEE Advanced 2021 — Paper 1 — Question 45

Let α,β\alpha, \beta and γ\gamma be real numbers such that the system of linear equation x+2y+3z=α,4x+5y+6z=β,7x+8y+9z=γ−1\begin{gathered}x+2 y+3 z=\alpha ,4 x+5 y+6 z=\beta ,7 x+8 y+9 z=\gamma-1\end{gathered} is consistent. Let ∣M∣|\mathrm{M}| represent the determinant of the matrix M=[α2γβ10−101]M=\left[\begin{array}{ccc}\alpha & 2 & \gamma \\\beta & 1 & 0 \\-1 & 0 & 1\end{array}\right] Let P be the plane containing all those (α,β,γ)(\alpha, \beta, \gamma) for which the above system of linear equations is consistent, and DD be the square of the distance of the point (0,1,0)(0,1,0) from the plane PP.

The value of ∣M∣\left| \text{M} \right| is

Answer: 1

Numerical answer — enter this value.

Step-by-step solution

∣M∣=∣α2γβ10−101∣(C1→C1+C3)=∣α+γ2γβ10001∣(R1→R1−2R2)\left| M \right|=\left| \begin{matrix}\alpha & 2 & \gamma \\\beta & 1 & 0 \\-1 & 0 & 1 \\\end{matrix} \right|\left( {{C}_{1}}\to {{C}_{1}}+{{C}_{3}} \right)=\left| \begin{matrix}\alpha +\gamma & 2 & \gamma \\\beta & 1 & 0 \\0 & 0 & 1 \\\end{matrix} \right|\left( {{R}_{1}}\to {{R}_{1}}-2{{R}_{2}} \right) \left| \begin{matrix}\alpha +\gamma -2\beta & 0 & \gamma \\\beta & 1 & 0 \\0 & 0 & 1 \\\end{matrix} \right|=\left| \begin{array}{*{35}{l}}1 & 0 & \gamma \\\beta & 1 & 0 \\0 & 0 & 1 \\\end{array} \right|=1

Answer key and solution verified before publishing.

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Exam
JEE Advanced 2021
Paper
Paper 1
Subject
Mathematics
Chapter
Straight lines
Topic
Locus