Question

If \(a,b,c\) are the intercepts made by the plane passing through the point \((1,2,3)\) parallel to the plane \(3x+4y-5z=0\) on \(X,Y,Z\)-axes respectively, then \(3a+b+5c=\)

• A. \(0\)
• B. \(1\)
• C. \(-1\)
• D. \(2\)

Hint:

Planes parallel to each other have the same coefficients of \(x,y,z\). Only the constant term changes.

Answer 1

The Correct Option is C

Step 1: Write the equation of any plane parallel to $ 3x+4y-5z=0 $.
Parallel planes share the same normal, so the required plane is $ 3x+4y-5z=k $.
Step 2: Substitute the given point (1,2,3) to find k.
\[ 3(1)+4(2)-5(3) = 3+8-15 = -4 \implies k = -4 \] The plane is $ 3x+4y-5z+4=0 $.
Step 3: Find the x-intercept a (set y=0, z=0).
$ 3a = -4 \implies a = -\frac{4}{3} $.
Step 4: Find the y-intercept b (set x=0, z=0).
$ 4b = -4 \implies b = -1 $.
Step 5: Find the z-intercept c (set x=0, y=0).
$ -5c = -4 \implies c = \frac{4}{5} $.
Step 6: Compute $ 3a+b+5c $.
\[ 3\!\left(-\frac{4}{3}\right)+(-1)+5\!\left(\frac{4}{5}\right) = -4-1+4 = -1 \] \[ \boxed{-1} \]