Statement-1 : The plane 5x + 2z - 8 = 0 contains the line 2x - y + z - 3= 0 and 3x + y + z = 5, and is perpendicular to 2x - y - 5z - 3 = 0.
Statement-2 : The plane 3x + y + z = 5 meets the line x-1 = y + 1 = z - 1 at the point (1, 1, 1).
Answer & explanation
Correct answer: option 3
The equation of the family of planes containing the line 2x - y + z - 3 = 0, 3x + y + z = 5 is
$2x - y + z - 3 + λ (3x + y + z - 5 ) = 0 $
For $ λ = 1,$ this reduces to 5x + 2z - 8 = 0
So, the plane 5x + 2z - 8 = 0 contains the given line.
Also, $ 2 × 5 - 1 × 0 - 5 ×2 = 0 $
So, the plane 5x + 2z - 8 = 0 is perpendicular to
2x - y - 5z - 3 = 0.
Hence, statement-1 is true.
The coordinates of any point on line $\frac{x-1}{1}=\frac{y+1}{1}=\frac{z-1}{1}$ are (r + 1 , r-1, r+1).
If this point lies on the plane 3x + y + z = 5. Then,
$3r + 3 + r - 1 + r + 1 = 5 ⇒ r = \frac{2}{5}$
Thus, the line meets the plane at $\left(\frac{7}{5}, -\frac{3}{5}, \frac{7}{5}\right)$.
So, statement-2 is not true.