Can (a ==1 && a== 2 && a==3) ever evaluate to true?

I couldn't resist - the other answers are undoubtedly true, but you really can't walk past the following code:

var aﾠ = 1;
var a = 2;
var ﾠa = 3;
if(aﾠ==1 && a== 2 &&ﾠa==3) {
    console.log("Why hello there!")
}

Note the weird spacing in the if statement (that I copied from your question). It is the half-width Hangul (that's Korean for those not familiar) which is an Unicode space character that is not interpreted by ECMA script as a space character - this means that it is a valid character for an identifier. Therefore there are three completely different variables, one with the Hangul after the a, one with it before and the last one with just a.

Check out the validation on Mathias' variable name validator. If that weird spacing was actually included in their question, I feel sure that it's a hint for this kind of answer.

Don't do this. Seriously.

IT IS POSSIBLE!

var i = 0;

with({
  get a() {
    return ++i;
  }
}) {
  if (a == 1 && a == 2 && a == 3)
    console.log("wohoo");
}

This uses a getter inside of a with statement to let a evaluate to three different values.

If it is asked if it is possible (not MUST), it can ask "a" to return a random number, it would be true if it generates 1,2 and 3 sequentially.

with({
  get a() {
    return Math.floor(Math.random()*4);
  }
}){
  for(var i=0;i<1000;i++){
    if (a == 1 && a == 2 && a == 3){
      console.log("after "+(i+1)+" trials, it becomes true finally!!!");
      break;
    }
  }
}

It can be accomplished using the following in the global scope. For nodejs use global instead of window in the code below.

var val = 0;
Object.defineProperty(window, 'a', {
  get: function() {
    return ++val;
  }
});
if (a == 1 && a == 2 && a == 3) {
  console.log('yay');
}

This answer abuses the implicit variables provided by the global scope in the execution context by defining a getter to retrieve the variable.

This is also possible using a series of self-overwriting getters:

(This is similar to jontro's solution, but doesn't require a counter variable.)

(() => {
	"use strict";
	Object.defineProperty(this, "a", {
		"get": () => {
			Object.defineProperty(this, "a", {
				"get": () => {
					Object.defineProperty(this, "a", {
						"get": () => {
							return 3;
						}
					});
					return 2;
				},
				configurable: true
			});
			return 1;
		},
		configurable: true
	});
	if (a == 1 && a == 2 && a == 3) {
		document.body.append("Yes, it’s possible.");
	}
})();

When you can't do anything without regular expressions:

var a = {
  r: /\d/g, 
  valueOf: function(){
    return this.r.exec(123)[0]
  }
}

if (a == 1 && a == 2 && a == 3) {
    console.log("!")
}

I don't see this answer already posted, so I'll throw this one into the mix too. This is similar to Jeff's answer with the half-width Hangul space.

var a = 1;
var ａ = 2;
var а = 3;
if(a == 1 && ａ == 2 && а == 3) {
    console.log("Why hello there!")
}

You might notice a slight discrepancy with the second one, but the first and third are identical to the naked eye. All 3 are distinct characters:

a - Latin lower case A
ａ - Full Width Latin lower case A
а - Cyrillic lower case A

The generic term for this is "homoglyphs": different unicode characters that look the same. Typically hard to get three that are utterly indistinguishable, but in some cases you can get lucky. A, Α, А, and Ꭺ would work better (Latin-A, Greek Alpha, Cyrillic-A, and Cherokee-A respectively; unfortunately the Greek and Cherokee lower-case letters are too different from the Latin a: α,ꭺ, and so doesn't help with the above snippet).

There's an entire class of Homoglyph Attacks out there, most commonly in fake domain names (eg. wikipediа.org (Cyrillic) vs wikipedia.org (Latin)), but it can show up in code as well; typically referred to as being underhanded (as mentioned in a comment, [underhanded] questions are now off-topic on PCCG, but used to be a type of challenge where these sorts of things would show up). I used this website to find the homoglyphs used for this answer.

Alternatively, you could use a class for it and an instance for the check.

function A(value) {
    value = value || 0;
    this.valueOf = function () { return ++value; };
}

var a = new A;

if (a == 1 && a == 2 && a == 3) {
    console.log('bingo!');
}

Rule number one of interviews; never say impossible.

No need for hidden character trickery.

window.__defineGetter__( 'a', function(){
        if(typeof(i) !== 'number'){
            i = 0;
        }
        return ++i;
    });
    
    if( a == 1 && a == 2 && a == 3 ){
        alert('Oh dear, what have we done?');
    }

Honestly though, whether there is a way for it to evaluate to true or not (and as others have shown, there are multiple ways), the answer I'd be looking for, speaking as someone who has conducted hundreds of interviews, would be something along the lines of:

"Well, maybe yes under some weird set of circumstances that aren't immediately obvious to me... but if I encountered this in real code then I would use common debugging techniques to figure out how and why it was doing what it was doing and then immediately refactor the code to avoid that situation... but more importantly: I would absolutely NEVER write that code in the first place because that is the very definition of convoluted code, and I strive to never write convoluted code".

I guess some interviewers would take offense to having what is obviously meant to be a very tricky question called out, but I don't mind developers who have an opinion, especially when they can back it up with reasoned thought and can dovetail my question into a meaningful statement about themselves.

Actually the answer to the first part of the question is "Yes" in every programming language. For example, this is in the case of C/C++:

#define a   (b++)
int b = 1;
if (a ==1 && a== 2 && a==3) {
    std::cout << "Yes, it's possible!" << std::endl;
} else {
    std::cout << "it's impossible!" << std::endl;
}

I'll leave as homework the demonstration for others languages. Have fun.

Here's another variation, using an array to pop off whatever values you want.

const a = {
  n: [3,2,1],
  toString: function () {
    return a.n.pop();
  }
}

if(a == 1 && a == 2 && a == 3) {
  console.log('Yes');
}

Okay, another hack with generators:

const value = function* () {
  let i = 0;
  while(true) yield ++i;
}();

Object.defineProperty(this, 'a', {
  get() {
    return value.next().value;
  }
});

if (a === 1 && a === 2 && a === 3) {
  console.log('yo!');
}

This one uses the defineProperty with a nice side-effect causing global variable!

var _a = 1

Object.defineProperty(this, "a", {
  "get": () => {
    return _a++;
  },
  configurable: true
});

console.log(a)
console.log(a)
console.log(a)

Even without confusing naming, overloading, or random variables, a == 1 && a == 2 && a == 3 can return true in multi-threaded environments as the value of a may change between each comparison as long as it is not thread-safe.

if you are allowed to reverse numbers and variables, the trick can be done with valueOf:

var i = 0
var a = {i: 0, valueOf: () => ++ i}
1 == a && 2 == a && 3 == a
// => true